\chapter{Stability}
\section {Quadratic Action}
{\bf Definition 1:}
$G$ is \emph{$\pi$-separable} iff every composition factor of $G$ is
either a $\pi$-group or a $\pi'$ group.
$G$ is \emph{$\pi$-solvable} iff every composition factor of $G$ is
either a solvable $\pi$-group or a $\pi'$ group.
\\
\\
{\bf Theorem 1:}
(1) If $G$ is $\pi$ separable iff the upper and lower $\pi$ series of $G$ terminate at
$G$.\\
(2) If $G$ is $\pi$-separable ($\pi$-solvable) so are subgroups and homomorphic images.
$G$.\\
(3) If $G$ is $\pi$-separable a minimal normal subgroup of $G$ is either a
$\pi$-group or a
$\pi'$-group.
\begin{quote}
\emph{Proof:}
For 3,
let $K$ be a minimal normal subgroup of the $\pi$ separable group $G$.
$K$ is characteristically simple an is thus a direct product of simple groups.
There is a composition series in which one of the isomophic subgroups in the
direct product is the last term and this must be a $\pi$ or $\pi'$ group.
For 2, homomorphic and normal subgroups are sutomatically $\pi$-separable (or $\pi$-solvable).
Let $H \le G$ and $K$ be a minimal normal subgroup of $G$, ${\overline G}= G/K$.
By induction, ${\overline H}$ is $\pi$-separable so we only need to show $H \cap K$ is;
this follows from 3 since $K$ is either a $\pi'$ group or a solvable $\pi$ group.
For 1, if the upper or lower series terminate, they can be refined to a composition series and
each of the factors is either a $\pi$ group or a $\pi'$ group.
In either case, $G$ is $\pi$ separable.
Conversely, if $G$ is $\pi$-separable and the upper $\pi$ series of $G$ terminates in
the proper subgroup $H$ of $G$, putting ${\overline G}= G/H$, we have
$O_{\pi}(G) = O_{\pi'}(G)= 1 $. But ${\overline G}$ is $\pi$ separable by 2 and so a minimal
normal subgroup ${\overline K}$ is either a $\pi$ or $\pi'$ group by 3 and thus either
${\overline K} \subseteq O_{\pi}({\overline G})$ or
${\overline K} \subseteq O_{\pi'}({\overline G})$ and either is a contradiction establishing 1.
\end{quote}
{\bf Theorem 2:}
If $G$ is $\pi$ separable and ${\overline G}= G/O_{\pi'}(G)$ then
$C_{\overline G}(O_{\pi}({\overline G})) \subseteq O_{\pi}({\overline G})$.
\begin{quote}
\emph{Proof:}
STS this when
$O_{\pi'}( G ) = 1$.  Set $H= O_{\pi}(G)$ and $C= C_G(H)$ so $C \cap Z = {\mathbb Z}(H)$ and
we must show $C= {\mathbb Z}(H)$.  $O_{\pi}(C) \thinspace char \thinspace \lhd G$ so
$O_{\pi}(C) \lhd G$ and hence
$O_{\pi}(C) \subseteq H$.  Thus $O_{\pi}(C)= C \cap H = {\mathbb Z}(H)$.  On the other hand,
${\mathbb Z}(H) \lhd G$ and
${\mathbb Z}(H) \subseteq C$ so
${\mathbb Z}(H) \subseteq O_{\pi}(C)$ and thus
${\mathbb Z}(H) = O_{\pi}(C)$.
Assume, by way of contradiction, that $C \supset {\mathbb Z}(H)$ then
$C \supset O_{\pi}(C)$.  $C$ is $\pi$-separable so
$L= O_{\pi, \pi'}(C) \subset O_{\pi}(C)$.  $L/O_{\pi}(C)$ is a $\pi'$ group,
${\mathbb Z}(H)= O_{\pi}(C)$ is a normal $S_p$ subgroup of $L$ and by Schur-Zassenhaus,
${\mathbb Z}(H)$ has a normal complement, $K \ne 1$ in $L$ which is a normal $S_{\pi'}$
subgroup of $L$.  But $K \subseteq C$ and $[C, {\mathbb Z}(H)]=1$ so
$L= {\mathbb Z}(H) \times K$ and since $K$ is a $\pi'$ group, $K \lhd G$ and
$K \subseteq O_{\pi'}(G)= 1$, a contradiction.
\end{quote}
{\bf Theorem 3:}
If $G$ is $\pi$-solvable
$C_G(P \cap O_{p', p}(G)) \subseteq O_{p', p}(G)$.
\begin{quote}
\emph{Proof:}
Let ${\overline G} = G/O_{p'}(G)$.  $C_{\overline G}({\overline P}) \subseteq  O_{p}({\overline G})$.
By coprime action, $C_{\overline G}({\overline P}) = C_G(P) O_{p'}(G)/O_{p'}(G)$.  By Theorem 2,
$C_G(O_{p',p}(G)) \subseteq O_{p',p}(G)$ and the result follows.
\end{quote}
{\bf Theorem 4:}
Let $G$ be a $p$-solvable group in which $O_{p'}(G)= 1$ and $H= O_p(G)$, then
$G/H$ is faithfully represented on $H/\Phi(H)$.
\begin{quote}
\emph{Proof:}
Let $g$ act by conjugation on $H$.
If $[g, H]=1$, $g \in p(G)$ because $O_{p'}(G)=1$.  So $gH$ acts on $H$ and is a $p'$-element.
By the result in the critical subgroups section, if $gH$ acts non-trivially on $H$ then
$gH$ acts non-trivially on $H/\Phi(H)$.
\end{quote}
{\bf Definitions 2:} If $V$ is an elementary abelian $p$-group then $a$
acts quadratically on $V$ if $[V,a,a]=1$, in which  case, $v^{(a-1)^2}=0$.
$G$ is said to be
\emph{$p$-separable} if two non conjugate elements of $G$ remain non-conjugate in
some finite $p$-group endomorphic image of $G$.
The \emph{upper $\pi$ series}  is
$\{ 1 \} \subseteq O_{\pi}(G) \subseteq O_{\pi, \pi'}(G) \subseteq O_{\pi, \pi', \pi}(G) \ldots$
The \emph{lower $\pi$ series} is
$\{ 1 \} \subseteq O_{\pi'}(G) \subseteq O_{\pi', \pi}(G) \subseteq O_{\pi', \pi, \pi'}(G) \ldots$.
\\
\\
{\bf Examples of quadratic action:} \\
(a) $G$ a $p$-group acting on the elementary abelian $p$-group $V$: $|V/C_V(G)|= p$;\\
(b) $G \in S_p(SL(V))$ acting on $V$ the vector space over $F_{p^m}$; in this case, note $x \in S_p(SL_2(p)) \rightarrow (x-1)^2 = 0$.\\
(c) $V, G \lhd H$, $G$-abelian, since $[V,G] \subseteq V \cap G$ and $[V \cap G, G] = 1$.
\\
\\
{\bf Theorem 5:}
If $G$ acts quadratically on $V$ then (a) $[v^n,a]=[v,a^n]=[v,a]^n$,
(b) $|V| \le |C_V(a)|^2$, (c) $G/C_G(V)$ is an elementary abelian $p-$group.
\begin{quote}
\emph{Proof:}
$[v,a^2]= [v,a] [v,a]^a$ but quadratic action gives $[v,a]^a= [v,a]$ so
$[v,a^2]= [v,a]^2$.  $[V,G']= 1$ so $G' \subseteq C_G(V)$ and so $G/C_G(V))$ is abelian.
Since $[v, a^p] = 1$, it is elementary abelian.  For (b), note that
$V/C_V(G) \cong [V, a] \le C_V(a)$.
\end{quote}
{\bf Theorem 6:}
Let $G$ act on an $F_q$ vector space $W \ne 0$, $q=p^m$.  Suppose $G= \langle a,b \rangle $ and
$a, b$ act quadratically on $W$, $G/C_G(W)$ is not a $p-$group, $|ab|=p^ek, k \mid (p-1)$
then $\exists \varphi: G \rightarrow SL_2(q)$.
\begin{quote}
\emph{Proof:}
By induction on $|G|+ dim(W)$.  If action is not faithful, we're done by induction.
Let $W_1$ be a maximal $G$-invariant subspace of $W$.  If $G/C_G(W_1)$ is not a
$p$-group,then $C_G(W/W_1)$ is not a $p$-group; if $W_1 \ne 0$, again we're done by induction.
So $G$ acts faithfully and irreducibly on $W$ and if $a, b$
are $p$-elements, they act quadratically and $G$ is not abelian.
By Schur, $< \langle ab \rangle $ acts as a scalar on a minimal
$ \langle ab \rangle $-invariant subspace
of $W$.  $\exists 0 \ne W \in W, \lambda \in F_q^*: w^{ab}= \lambda w$.
If $w^a \in F_q w$, it is $G$-invariant and $W= F_qw$ by irreducibility and $G$ is
abelian.  Contradiction.  Thus $W_1 = F_q w + F_q w^a$ is $2$-dimensional and we
have:
$[w,a] \in C_{W_1}(a)$ and
$w^{b^{-1}} -w \in C_{W_1}(b)$.  So $(w^a)^a \in W_1$, $w^b \in W_1$ and $w^{ab} \in W_1$.
so $W_1$  is $G$-invariant and $W= W_1$.  $G \le SL(W)$ since $SL_2(q)$ is generated by
$p$-elements $a, b$.
\end{quote}
{\bf Definition 3:} $G$ is \emph {$p$-stable} if $\forall a \in G, [V,a,a]=1$ implies
$a C_G(V) \in O_p ( G/C_G(V))$.
\\
\\
{\bf Question:}  If $V/F_q$, $q= p^n$, is a faithful $G$-module, when does a $p$ element of
$G$ have a quadratic minimal polynomial?  This is trivial for $p=1$ and is true for
all elements of $SL_2(q)$.  Let $G$ be a group and $O_p(G)=1, p \ne 2$
A faithful representation of $G$ on $V$, $\varphi$ is \emph{$p$-stable} if
no $p$-element of $\varphi(G)$ has a quadratic minimal polynomial.  $G$ is $p$-stable
if all such faithful representations of $G$ are $p$-stable.
\\
\\
{\bf Lemma:} If $G$ is $p$-stable and $a$ acts trivially on $[V,a]$ then $ \langle A \rangle
C_G(V)/C_G(V)$
is a $p$-group.
\begin{quote}
\emph{Proof:}
$[V,a,a]=1$ so $a C_G(v)
\in O_p(G/C_G(V))$ and $\langle a C_G(V) \rangle = \langle A \rangle
\subseteq  O_p(G/C_G(V))$.
\end{quote}
{\bf Theorem 7:} Let $p \ne 2$ and $G$ be faithful on $V$.
Suppose (1) $G= \langle a,b \rangle $ where $a$ and $b$ act quadratically on $V$ and
(2) $G$ is not a $p-$group then (1) the Sylow $2$ subgroups of $G$ are not abelian and
(2) If $Q$ is a normal $p'$-subgroup of $G$ and $[Q,a] \ne 1$ then $p=3$ and there
is a section of $G$ isomorphic to $SL_2(3)$.  If $p \ne 2$.
\begin{quote}
\emph{Proof:}
Let $|ab|= p^e k$ and $q$ be a power of $p$ with $k \mid (q-1)$.  Write $V$ additively
as a vector space over $F_p$ and choose a basis $ \langle v_1 , v_2 , \ldots , v_n \rangle $;
the action
can be extended to an action of $G$ over $W$.  By the previous result,
$\exists \varphi: G \rightarrow SL_2(q)$ with
$G^{\varphi}= \langle a^{\varphi}, b^{\varphi} \rangle $ and
$G$ is not a $p$-group so it is not $p$-closed.  This gives (a).  (b) follows since
if $Q^{\varphi}$ is an $a^{\varphi}$-invariant $p'$-subgroup
such that $[Q^{\varphi}, a^{\varphi}] \ne 1$.
\end{quote}
{\bf Theorem 8:}
Suppose $p \ne 2$ and the action of
$G$ on $V$ is faithful and not $p-$stable then (1) the Sylow $2$-subgroups of
$G$ are non-Abelian and (2) if $G$ is $p-$separable
then $p=3$ and there is a section of
$G$ isomorphic to $SL_2(3)$.
\begin{quote}
\emph{Proof:}
$\exists a \in G \setminus O_p(G)$ such that $[V,a,a] = 1$.  Let ${\cal K}$ be the
$G$-composition factors of $V$.  $O_p(G)= \bigcap_{W \in {\cal K}} C_G(W)$.  Hence,
$\exists W \in {\cal K}, a \notin C_G(W)$ so a $C_G(W) \notin O_p(G/C_G(W)) = 1$.
Thus $G/C_G(W)$ and $W$ satisfy the hypothesis and we can assume $W=V, O_p(G)=1$ by
induction.  By Baer, $\exists b \in a^G: G_1 = \langle a,b \rangle $ is not a $p$-group.  Now
(a) follows from the previous result.  If $G$ is $p$-seperable, put $Q= O_{p'}(G)$ then
$[Q, a] \ne 1$ and $b$ can be chosen in $a^Q$ so we get (b).
\end{quote}
{\bf Theorem 9:} Suppose $G$ acts faithfully on $V$ and $E_1 , E_2$ are
two subnormal subgroups of $G$
such that $[V,E_1 , E_2 ]=1$ then $[E_1 , E_2 ] \le O_p (G)$.
\begin{quote}
\emph{Proof:}
By hypothesis, $V_1= [V, E_1]$ is invariant under $E= \langle E_1 , E_2 \rangle $ so
$E_0 = C_E(V_1)$ and $E^0= C_E(V/V_1)$ are normal in $E$.
$E_0 \cap E^0$ acts quadratically on $V$ and is a $p$-group by the earlier result.
Thus $E_0 \cap E^0 \le O_p(E)$.
$E_1 \le E^0$ and
$E_2 \le E_0$ so $[E_1 , E_2 ] \le [E^0 , E_0] \le E^0 \cap E_0 \le O_p(E)$ so
$E$ and $O_p(E)$ are subnormal in $G$ and hence, $O_p(E) \le O_p(G)$.
\end{quote}
$Q_8= \langle
\left(
\begin{array}{cc}
i & 0 \\
0 & -i \\
\end{array}
\right),
\left(
\begin{array}{cc}
0 & -1 \\
-1 & 0 \\
\end{array}
\right) \rangle $.
\section {Replacement Results}
{\bf Observation:} $A^* = C_A([V,A])$ acts quadratically on $V$.
\\
\\
{\bf Condition ${\cal Q}_1$:}
$|A| |C_V(A)| \ge |A^*| C_V(A^*)|, \forall A, A^*$.
\\
\\
{\bf Condition ${\cal Q}_2$:}
$A/C_A(V)$ is an elementary abelian $p$-group.
\\
\\
{\bf Definition 4:}  ${\cal A}_V(G) = \{ A \le G: A$ satisfies ${\cal Q}_1$ and ${\cal Q}_2 \}$.
\\
\\
{\bf Theorem 10:}
Suppose $A$ acts on an elementary abelian $p$-group, $V$, with $A/C_A(V)$ abelian.
Let $U \le V$.  Then $\exists A^* \le A$ such that one of the following holds:
(a) $|A| |C_V(A)| \l |A^*| |C_V(A^*)|$, or
(b)  $A^* = C_A([U,A]), C_V(A^*)= [U,A] C_V(A), |A| |C_V(A)| = |A^*| |C_V(A^*)|$.
\begin{quote}
\emph{Proof:}
Assume ${\cal Q}_1$ does not hold.   $\forall B \le A$, so $|A| |C_V(A)| \ge |B| |C_V(B)|$.
Put $A^* = C_A([U,A])$.  $[U, A, A^*]=1$ and $A/C_A(V)$ is abelian so
$[A, A^*, U] =1  \rightarrow [U,A^* A] = 1$ so $[U, A^*] \le C_V(A)$.
\\
\\
\emph{Claim:} $|A| |C_V(A)| \le |A^*| |[U,A] C_V(A)|$.
\\
\\
Assuming claim, $|A^*| |C_V(A^*)| \le |A| |C_V(A)| \le |A^*| |[U,A] C_V(A)| \le |A^*| |C_V(A^*)|$
and we're done.
\\
\\
\emph{Proof of claim:}
Let $Y= C_V(A), X= [U,A]$.
We can assume $|U|= p, U= \langle u \rangle $ and $[U, A]= [u,a]$.
Let $\varphi: A/A^* \rightarrow (XY)/Y$ by $a A^* \mapsto [u,a]Y$ is well defined.
$\forall a^* \in A^*$.
$[u, a^*, a]= [u,a] [u, a^*]^a \in [u,a]Y$.
If $\varphi$ is injective, $|A/A^*| \le (XY)/Y|$ and the result follows.
Let $a_1, a_2 \in A$ such that $[u, a_1]Y= [u, a_2 ]Y$.
$[u, a_1] [u, a_2 ]^{-1} \in Y$ then $[u, a_1 a_2^{-1}] \in Y$ so
$[u, A, A_1 s_2^{-1}]= 1 $
and $a_1 a_2^{-1} \in C_A([U,A])=A^*$.  Thus $\varphi$ is injective.
Now assume $|U| > p$, $|U:U_1|= p, U= U_1 \langle u \rangle $.
Put $X_1= [U_1, A]$, $A_1 = C_A(X_1)$ and
$X_2= [U_2, A]$, $A_2 = C_A(X_2)$.
Note $X_1 X_2 C_V(A)= X C_V(A), A^* = A_1 \cap A_2$ and
$X_1 C_V(A) \cap X_2 C_V(A) \le C_V(A_1 A_2 )$.
By induction on $|U|$,
$|A| |C_V(A)| = |A_i | |X_i:C_V(A)|$.  Hence,
$|A| |C_V(A)| \ge
|A_1 A_2| |C_V(A_1 A_2 ) | \ge
{\frac {|A_1| |A_2| |X_1 C_V(A)| |X_2 C_V(A)|}
{|A_1 \cap A_2| |X_1 C_V(A)| |X_2 C_V(A)|} }
= {\frac {|A|^2 |C_V(A)|^2} {|A^*| |X C_V(A)|}}$ and this proves the claim.
\end{quote}
{\bf Theorem 11:}
$A \in {\cal A}_V(G)$ and $A^* = C_A([V,A])$ then
$|A/A^*| = |C_V(A^*)/C_V(A)|$ and $C_V(A^*)= [V,A] C_V(A)$.
\begin{quote}
\emph{Proof:}
By the previous result, every quadratically acting subgroup, $A$, satisifes
${\cal Q}_2$ and the result follows from the second conclusion of that theorem, with $U=V$.
\end{quote}
{\bf Timmesfeld Replacement Theorem:}
Let $A \in {\cal A}_V(G)$ and $U \le V$ then $C_A([U,A]) \in {\cal A}_V(G)$ and
$C_V(C_A([U,A]))= [U,A] C_V(A)$.  Moreover,
$[V,C_A(([U,A])] \ne 1$ if $[V,A] \ne 1$.
\begin{quote}
\emph{Proof:}
Let  $A^* C_A([U,A])$.  Since $A \in {\cal A}_V(G)$, we can apply the previous
result so
$|A^*| |C_V(A^*)| = |A| |C_V(A)|$ and $C_V(A^*)= [U,A] C_V(A)$.
$\forall A_0 \le A, {\cal Q}_1$ gives
$|A_0| |C_V(A_0)| \le |A^*| |C_V(A^*)|$.
Hence
$A^* \in {\cal A}_V(G)$, we may assume $[V, A^*] =1$ then $V= [U,A] C_V(A)= [V,A] C_V(A)$.
In particular, $[V,A,A]=[V,A]$ but then $[V,A] = 1$ since $A/C_A(V)$ is a $p$-group.
\end{quote}
{\bf Definition 5:} ${\cal A}_V(G)_{min}= \{ A \in {\cal A}(G), [V,A] \ne 1 \}$.
\\
\\
{\bf Theorem 12:}  Every element of ${\cal A}_V(G)_{min}$ acts quadratically and non-trivially on
$V$.
\begin{quote}
\emph{Proof:}
Let $A \in
{\cal A}_V(G)_{min}$.  By previous result, $A^*= C_A([V,A])$ is also in
${\cal A}_V(G)$ and $[V, A^*] \ne 1$.  Minimality forces $A=A^*$ and thus
$[V,A,A]=1$.
\end{quote}
{\bf Theorem 13:}  Suppose $G$ is $p$-stable on $V$ and $O_p(G/C_G(V))=1$ then every element of
${\cal A}_V(G)_{min}$ acts trivially on $V$.
\begin{quote}
\emph{Proof:}
Follows from previous result.
\end{quote}
{\bf Theorem 14:}
Let $V= \langle C_V(S), S \in S_p(G) \rangle $ then $O_p(G/C_G(V))=1$.
\begin{quote}
\emph{Proof:}
Let $S \in S_p(G)$, $Z= C_V(S), C= C_G(V)$.  $V= \langle Z^G \rangle $.  Let $C \le D \le G$:
$D/C \cong O_p(G/C)$ then $D \cap S \in S_p(D)$ and $D= C(D \cap S)$ and by
Frattini: $G=C N_G(D \cap S)$.  So $V= \langle Z^{N_G(D \cap S)} \rangle $, so $[V, D \cap S]= 1$
and $D=C$.
\end{quote}
{\bf Theorem 15:}
Let $C_G(O_p(G)) \le O_p(G)$ then $V= \langle \Omega({\mathbb Z}(S)), S \in S_p(G) \rangle $
is an elementary
abelian normal subgroup of $G$ and $O_p(G/C_G(V))=1$.
\begin{quote}
\emph{Proof:}
Let $S \in S_p(G)$ then $\Omega ({\mathbb Z}(S)) \le C_G(O_p(G)) \le O_p(G) \le S$ so
$V$ is contained in $\Omega({\mathbb Z}(O_p(G)))$ and $\Omega({\mathbb Z}(S))= C_V(S)$.
Now the result follows from the previous result.
\end{quote}
{\bf Definition 6:} Let ${\cal E}(G)$ be the set of elementary elementary subgroups of $G$ and
$m$ the the size of the element of ${\cal E}(G)$ of maximal order.
$J(G)= \langle A \in {\cal A}(G): |A|=m \rangle $.
\\
\\
{\bf Theorem 16:}
Let $A \in {\cal A}(G)$ act quadratically
on $V$ and $A_0 = [V,A] C_A([V,A])$ then $A_0$ is in ${\cal A}(G)$ and acts quadratically
on $V$ and if $[V,A] \ne 1$ then $[V, A_0 ] \ne 1$.
\begin{quote}
\emph{Proof:}
Let $X= [V,A]$ and $A^*= C_A(X)$ with $A_0= A^*X$.  $A_0$ is elementary abelian
and $[V,A_0, A_0] \le [V, A, A_0 ] =1$.  It suffices to show $|A|= |A_0|$ to establish
$A_0 \in {\cal A}(G)$.  The maximality of $A$ gives $C_V(A)= V \cap A= V \cap A^*$ and since
$X \cap A = X \cap A^*$, $|A| |A \cap V| = |A| |C_V(A)| = |A^*| |XC_V(A)|$.
and so
$|A| = {\frac {|A^*| |XC_V(A)|} {|C_V(A)|}} =
{\frac {|A^*|} {|X \cap C_V(A)|}} = {\frac {|A^*| |X|} {|X \cap A^*|}} = |A_0|$.
\end{quote}
{\bf Theorem 17:}
(a) ${\cal A}(G) \subseteq {\cal A}_V(G)$ and (b) $V \nleq {\mathbb Z}(J(G))$ then
$\exists A \in {\cal A}(G): [V,A] \ne 1$.
\begin{quote}
\emph{Proof:}
Let $ A^* \in {\cal A}(G)$ then $A^* C_V(A^*) \in {\cal E}(G)$ so
$|A| \ge |A^* C_V(A^*)|=
{\frac {|A^*| |C_V(A^*)|} {|A^* \cap V|}} \ge
{\frac {|A^*| |C_V(A^*)|} {|C_V(A)|}} $ and (a) follows.  (b) is clear.
\end{quote}
{\bf Condition ${\cal Q}_1'$:} $|A/C_A(V)| \ge |V/C_V(A)|$.
\\
\\
{\bf Theorem 18:}
Let ${\cal B}$ be the set of subgroups $A \le G$ satisfying
${\cal Q}_1'$ and
${\cal Q}_2$.  Let $A \in {\cal B}$ and suppose that $\forall A^* \le A, A^* \in {\cal B}$,
$|A^*/C_{A^*}(V)| |C_V(A^*)| \le |A/C_A(V)| |C_V(A)|$ then $A \in {\cal A}_V(G)$.
\begin{quote}
\emph{Proof:}
We need to verify ${\cal Q}_1$ for $A$.  Let $A^* \le A$.  If $A^*$ does not satisfy
${\cal Q}_1'$ then $A^* \notin {\cal B}$ and
$|A^*/C_{A^*}(V)| |C_V(A^*)| <
|V| \le |A/C_A(V)| |C_V(A)| $.  So
$|A/C_A(V)| |C_V(A)| \ge |A^*/C_{A^*}(V)| |C_V(A^*)|=
|(A^*/C_{A^*}(V))/C_A(V)| |C_V(A^*)|$.  The inequality also holds for $A^* \in {\cal B}$ since
the inequality holds.
Thus $\forall A^* \le A$:
$|A^*| |C_V(A^*)| \le |A^* C_{A}(V)| |C_V(A^*)| \le |A| |C_V(A)|$ and
$A$ satisfies ${\cal Q}_1$.
Assume $\exists A \in {\cal B}$ that act non-trivially on $V$.
Among all such choose it with the property that $|A/C_A(V)| |C_V(A)|$ is
maximal. Then the inequality in the theorem holds for $A$.  Thus
$A \in {\cal A}_V(G)$ and ${\cal A}(G)_{min} \ne \emptyset$.  A previous result insures
the existence that act quadratically and non-trivially on $V$.
\end{quote}
{\bf Theorem 19:} If ${\cal K} \in  \{N, S, \Pi\}$ then
$\forall G: O_{\cal K}(G)= \langle A: A \in {\cal K}, A \lhd \lhd G \rangle $.
\begin{quote}
\emph{Proof:}
Tor $A \lhd G$, this is clear.  May assume $A$ is not normal in $G$ so $\exists N \lhd G$ with
$A \lhd \lhd N<G$.  By induction,
$A \leq O_{\cal K}(N)$.
$O_{\cal K}(N) \lhd G$ and  ${\cal K} \in \{N , S , \Pi \}$.  Hence $A \leq O_{\cal K}(N) \leq O_{\cal K}(G)$ which
proves the result.
\end{quote}
\section{Stability and $SL_2(p)$}
{\bf Hall's remark on Thompson:}  If $P \in S_p(K)$ and $K \ne XP$ for any $X \in O_{p'}(K)$, then
there is a characteristic subgroup $D$ of $P$ of nilpotence class at most $2$ such that
$N_K(D)/C_K(D)$ is not a $p$-group.
\\
\\
{\bf Definition 7:}  Let $G$ be a $p$-constrained group with $O_{p'}(G)=1$.
$G$ is $p$\emph{-stable} if $\forall H \lhd G, H \in p(G)$,
$[H,x,x]=1 \rightarrow {\overline x} \in O_p(G/C_G(H))$.
\\
\\
{\bf Theorem 20:}
A group is $p$-separable iff
$H<G$ has a non-trivial $\pi$-closed factor group or, equivalently
$G$ has a normal series $1= A_0 < A_1 < \ldots <A_n=G$ of characteristic
subgroups and $A_i/A_{i-1}$ is a $\pi$ or $\pi'$ group.
\begin{quote}
\emph{Proof:}  Sort of a definition.
\end{quote}
{\bf Theorem 21:}
If $p \ne 2$, $O_{p'}(G)=1$ and $G$ is $p$-constrained or $p$-solvable and $SL_2(p)$ is
not involved in $G$, then $G$ is $p$-stable.
\begin{quote}
\emph{Proof:}
Suppose $H \lhd G$ is a $p$-group.  $G/C_G(H)$ acts on $H/\Phi(H)$.  Let $K=ker(\varphi)$
where $\varphi: G/C_G(H) \rightarrow Aut(H/\Phi(H))$.
\\
\\
\emph{Claim:} $K$ is a $p$-group.
\\
\emph{Proof of claim:} It suffices to show that if
${\mathbb Z}_q < G/C_G(H)$ acts non-trivially on $H$, it acts nontrivially on
$H/\Phi(H)$.  If ${\mathbb Z}_q$ acts trivially on $H/\Phi(H)$.  For each coset,
$\Phi(H)x$, by counting, at least one element of the coset is fixed, say $x_i$.
By the Burnside Basis Theorem, $ \langle x_i \rangle_i = H$.
Thus there is no $q$-element in $K$ and
$|K|=p^m$.
\\
\\
Put $L= G/C_G(H)/K$.  $L$ act faithfully on $H/\Phi(H)$.  Since $SL_2(p)$ is not involved,
$K$ is $p$ stable.  Now let $x \in G$ with $[H,x,x]=1$ and let the canonical map
$G/C_G(H) \rightarrow L$ be denoted by $\tilde {}$.
$[H,x,x]=1 \rightarrow [H,x,x] \in \Phi(H) \rightarrow
(\tilde {\overline x}-1)^2=0$, so
$\tilde {\overline x} \in O_p(\tilde{G/C_G(H)})$ we have
$\tilde {\overline x} \in O_p(G/C_G(H))$.
\end{quote}
{\bf Theorem 21:}
Let $G$ be a group with no non-trivial normal $p$-subgroup, $p \ne 2$ which satisfies one of
the following:
(1) $G$ has odd order;
(2) $G$ has an abelian Sylow $2$-subgroup;
(3) $G$ has a dihedral Sylow $2$-subgroup;
(4) $G \cong PSL_2(q) = L_2(q)$;
(5) $G$ is solvable and $p \ge 5$ or $p=3$ and $SL_2(3)$ is not involved in $G$
then $G$ is $p$-stable.
\begin{quote}
\emph{Proof:}
Generalization of earlier result.
\end{quote}
\section{The Thompson Subgroup}
{\bf Definition} Let ${\cal E}(G)$ be the elementary abelian subgroups of $G$ of maximal order.
$J(G)= \langle {\cal E}(G) \rangle)$.
\\
\\
{\bf $GL$ Lemma:} Let $G= GL_2(p)$, $p \neq 2$, $P \in S_p(G)$.  Suppose $L \in p'(G)$ and $P \subseteq N_G(L)$ and
if $S \in S_2(G)$, $S'=1$.  Then $P \subseteq C_G(L)$.
\begin{quote}
\emph{Proof:}   Sublemma: If $q$ is odd, $-I$ is the unique involution in $SL_2(q)$.  Let $P$ be
a $p$-group with at most one group of order $p$.  Either $P$ is cyclic or $p=2$ and
$P$ is generalized quaternion.
\\
By induction on $|L|$, we can assume $P$ centralizes every proper subgroup that it stabilizes.
Choose $q \mid |L:C_L(P)|$.  We can find a $P$ invariant Sylow $q$ subgroup $Q \subseteq L$.
Since $Q \neq C_L(P)$, $Q=L$.  So $L$ is a $q$-group.  $[L, P] \subseteq P$.
If $[L,P] < L$, $[L,P,P]=1$ and we're done.  So $[L,P]=P$, $L \subseteq G' \subseteq SL_2(p)$ since
$GL_2(p)/SL_2(p)$ is abelian.  If $q = 2$, $L$ is abelian and has a unique involution.
$L$ is a cyclic $2$-group and so is $Aut(L)$  $P$ cannot act non-trivially on $L$ so $q \neq 2$.
$|L| \mid (p-1)p(p+1)$ so $q \mid p-1$ or $q \mid p+1$. This $|L| \leq p+1$.  If $P$ acts non-trivially on $L$
then there must be a $p$-orbit of $L$ of size at least $p$. $|L|=p+1$ and $|L|$ is even but
$|L|$ is a power of $q$.  Contradiction.
\end{quote}
{\bf Normal P-Theorem:} Let $P \in S_p(G)$ and suppose (1) $G$ is $p$-solvable, (2) $p \ne 2$, (3) if
$R \in S_2(G)$, $R' = 1$, (4) $O_{p'}(G) = 1$, and (5) $P = C_G({\mathbb Z}(P))$ then $P \lhd G$.
\begin{quote}
\emph{Proof:}
Suppose $G$ is a minimal counter-example. $\exists Q \in S_p(G), P \neq Q: \langle P, Q \rangle G$.
$Q= P^g$ and $C_V(Q) = C_V(P)^g$.  Put $U= C_V(Q) \cap C_V(P)$. $|V:U| \leq |V:C_V(P)| |V:C_V(Q)| = p^2$ and
$U \lhd V$.  $G$ acts trivially on $U$: $[U, G] = 1$.  $G$ acts on $V/U$; let $K$ be the kernel of this map.
$[V,U] \subseteq U$ and $[V, K, K] =1$ so $K$ is a $p$-group.  Note $K \subseteq O_p(G)$ so
$K \subseteq P$ and $K \subseteq Q$. ${\overline G} = G/K$ has Sylow subgroups ${\overline P}, {\overline Q}$
and ${\overline G}$ acts faithfully on $V/U$ so $[{\overline P}, C_V(P)/U] = 1$.  ${\overline G}$ satisfies
all the hypothesis of the theorem so $K=1$ and $G$ acts faithfully on $V/U$.
Replace $V$ by $V/U$. $|V| \leq p^2$. $G \rightarrow Aut(V)$.  If $V$ is cyclic, $Aut(V)$ is abelian and
so is $G$ therefore $P \lhd G$.  So $V$ is elementary abelian and $Aut(V) = GL_2(p)$ and $O_p(G) = 1$ since
$|P| \leq p$ and $P$ is not normal.  Let $L = O_{p'}(G)$ and apply the previous lemma.  $[P, L]=1$ so by
Hall-Higman, $P \subseteq C_G(L) \subseteq L$ and $P=1$.
\end{quote}
{\bf Normal J-Theorem:} Let $P \in S_p(G)$ and suppose (1) $G$ is $p$-solvable, (2) $p \ne 2$, (3) if
$R \in S_2(G)$, $R' = 1$, (4) $G$ acts faithfully on some $p$-group, $V$ and (5) $|V:C_P(V)| \leq p$,
then $J(P) \lhd G$.
\begin{quote}
\emph{Proof:}
Let $G$ be a minimal counterexample.  $U=O_{p}(G) > 1$, ${\overline G} = G/U$,
${\overline L} = O_{p'}({\overline G})$, where $U \subseteq L$. \\
\\
\emph{Step 1:} (a) ${\mathbb Z}(P) \subseteq U$, (b) $U \subseteq H \subseteq G$ implies $O_{p'}(H)=1$ and
(c) $C_{{\overline G}}({\overline L}) \subseteq {\overline L}$.\\
\emph{Proof:}
Since $G$ is $p$-solvable and $O_{p'}(G)=1$,
by 1.2.3, $C_G(U) \subseteq O_p(G) = U$ but $U \subseteq P$ so
${\mathbb Z}(P) \subseteq C_G(U)$, proving (a).
Since $U = O_p(G)$ and $O_p({\overline G}) = 1$, so
$C_{{\overline G}}({\overline L}) \subseteq {\overline L}$ by 1.2.3,
proving (c).  For (b), let
$U \subseteq H \subseteq G$, and put $M= O_{p'}(H)$.  $M, U \lhd H$. $M \cap U =1$, since
$U$ is a $p$-group and $M$ is a $p'$-group, so $M \subseteq C_G(U) \subseteq U$ and thus $O_{p'}(H)=1$
and $M= M \cap U =1$ proving (b).
\\
\\
\emph{Step 2:} $\exists A \in {\cal E}(P): A \not\subseteq U$.\\
\emph{Proof:} If not, all members of ${\cal E}(P)$ are contained in $U$ and so $J(P) \subseteq U$.
By the $GL$ Lemma, $J(U) = J(P)$ is characteristic in $U$ and since $U \lhd G$, $J(P) \lhd G$,
which contradicts the fact that $G$ is a counterexample.
\\
\\
\emph{Step 3:} Let $UA \subseteq H \subset G$ and $H \cap P \in S_p(H)$ then ${\overline A}$ centralizes
${\overline {H \cap L}}$.\\
\emph{Proof:}  $H$ satisfies the first four hypothesis of the theorem.  A Sylow $2$-group of $H$ is abelian so
it meets condition (3).  $O_{p'}(H) = 1$ by 1(b).
Put $S = H \cap P \in S_p(H)$.  Since ${\mathbb Z}(P) \subseteq U \subseteq S \subseteq P$ by 1(a).
${\mathbb Z}(P) \subseteq {\mathbb Z}(S)$ and thus $C_H({\mathbb Z}(S)) \subseteq C_G({\mathbb Z}(P)) = P$.
So $C_H({\mathbb Z}(S)) = S$ is a $p$-group of $H$ containing the Sylow $p$ group $S$.
So $C_H({\mathbb Z}(S)) = S$ and $H$ satisfies the fifth hypothesis.  Since $H < G$, the theorem holds
for $H$ and so $J(S) \lhd H$.  Since $A \in {\cal E}(P)$ and $A \subseteq S \subseteq P$ and $A \in {\cal E}(S)$ so
$A \subseteq J(S)$.  Thus $[H \cap L, A] \subseteq [H \cap L, J(S)] \subseteq (H \cap L) \cap J(S) = L \cap J(S) \subseteq U$
(Reason: $H \cap L$ and $J(S)$ are normal in $H$ and $U$ is the unique Sylow $p$-subgroup of $L$).
$1 = [{\overline {H \cap L}}, {\overline {A}}]$.
\\
\\
\emph{Step 4:} $G = LA$, $P=UA$.\\
\emph{Proof:}  $H=LA$ and $UA \in p(H)$.  Further,  $|H:UA|= |L(UA):UA| = |L:L \cap UA|$ which divides the $p'$-number
$|L:Y|$.  So, $UA \in S_p(H)$, $UA = H \cap P$.
If $H < G$
then since $L \subseteq H$, step 3 gives
${\overline A} \subseteq C_{\overline G}({\overline L}) \subseteq {\overline L}$ (by step 1(c)).  Since ${\overline A}$ is
a $p$-group and ${\overline L}$ is a $p'$-group, ${\overline A} = 1$ and $A \subseteq U$.  This contradicts the choice of $A$
so $H = G$.  Finally, $UA = H \cap P = G \cap P = P$.
\\
\\
\emph{Proof:} Put $H=LA$, $UA \in p(H)$.  $|H:UA| = |LUA:UA|= |L:L \cap UA|$ so $UA \in S_p(H)$ and
$UA = H \cap P$.
so $UA \in S_p(H)$.  If $H \subseteq G$, since $L \subseteq H$, step 3 gives
${\overline A} \subseteq C_{{\overline G}}({\overline L}) \subseteq {\overline L}$.  Since
$A$ is a $p$-group and ${\overline L}$ is a $p'$-group, ${\overline A} =1$ and $A \subseteq U$.
This contradicts the choice of $A$ so $H=G$.  Finally, $UA = H \cap P = G \cap P = P$.
\\
\\
\emph{Step 5:} $|{\overline A}| = p$.\\
\emph{Proof:}
${\overline A} \ne 1$ since $A \not\subseteq U$.  ${\overline A}$ is elementary abelian, it STS ${\overline A}$ is cyclic.
${\overline A}$ acts coprimely on ${\overline L}$ and the action is faithful since $C_{\overline L}({\overline L}) \subseteq {\overline L}$
and ${\overline L} \cap {\overline A} = 1$.  A previous result shows ${\overline A}$ is cyclic so STS ${\overline A}$ acts trivially
on every ${\overline A}$-invariant proper subgroup of ${\overline L}$.
Suppose
${\overline M}$ is ${\overline A}$-invariant, ${\overline M} < {\overline L}$, we can assume
$U \subseteq M$. $A \subseteq N_G(M)$ so $MA$ is a group $P = UA \subseteq MA$.  Since $A$ is a $p$-group,
the $p'$ part of $|MA|$ is equal to the
$p'$ part of $|M|$ which is less than the $p'$-part of $|L|$ since
the $p'$ part of$|L:M| >1$.
It follows $MA < G$.  Apply step 3 to show ${\overline A}$ centralizes
${\overline {MA \cap L}} \supseteq M$, proving 5.
\\
\\
\emph{Step 6:}  Let $V= \{ z: z \in {\mathbb Z}(U) | z^p = 1\}$.  $V$ is an elementary abelian normal subgroup of $G$ so $G$ acts by
conjugation on $V$.  Since $V \subseteq {\mathbb Z} (U)$, the action by $U$ is trivial, so ${\overline G} = G/U$ on $V$.
Now we prove:
The action of ${\overline G}$ on $V$ is faithful. \\
\emph{Proof:}
Let $K = C_G(V)$ so ${\overline K}$ is the kernel of the action of ${\overline G}$ on $V$.
We argue $K$ is a $p$-group.
Let $Q \in S_q(K)$, $q \ne p$.  $Q$ acts coprimely on ${\mathbb Z}(U)$, so
and $Q$ fixes all elements of order $p$ in ${\mathbb Z}(U)$, these make up $V$.
$Q \subseteq K =C_G(V)$.   By Fitting, $Q$ acts trivially on ${\mathbb Z}(U)$ but
${\mathbb Z}(P) \subseteq U$ so ${\mathbb Z}(P) \subseteq {\mathbb Z}(U)$ and so $Q \subseteq C_G({\mathbb Z}(P)) = P$.
Thus $Q = 1$ and $K$ is a $p$-group, as claimed.   But $K \lhd G$.  So $K \subseteq O_p(G)=U$ and ${\overline K} = 1$ as needed.
\\
\\
\emph{Step 7:} $|V: V \cap A| \leq p$.\\
\emph{Proof:}  Put $D = U \cap A$ and $E = V \cap A$.  $|V:E| = |V:V \cap D| = |VD:D|$.  $D$ is an elementary abelian
subgroup of $U$ and $V$ is a central elementary abelian subgroup of $U$ and so $VD$ is elementary abelian.
Since ${\cal E}(P)$, $|VD| \leq |A|$ and so $|VD:D| \leq |A:D| = |{\overline A}| = p$.  We get $|V:E|= |VD:D| \leq p$
as required.
\\
\\
\emph{Step 8:} Contradiction. \\
\emph{Proof:}
We apply the Normal-P theorem to the action of ${\overline G}$ on $V$.
$|V: V \cap A| \leq p$. Now, $P=UA$ so ${\overline P} = {\overline A}$.
$[{\overline A}, V \cap A] = 1$ since $A' = 1$.
So $|V:C_V({\overline P})| \leq | V: V \cap A| \leq p$.  Now we can apply the Normal P theorem,
${\overline P} \lhd {\overline G}$ so $P \lhd G$ and $A \subseteq P \subseteq O_p(G)=U$, which is not
the case.
\end{quote}
{\bf Theorem 22:}
(1) $J(G) \; char \; G$ and $J(G) > 1$ if $p \in \pi(G)$;
(2) If $J(G) \le U \le G$ then $J(G) = J(U)$;
(3) $J(G)= \langle J(S): S \in S_p(G) \rangle $;
(4) If $x \in C_G(J(G))$ and $|x|=p$, then $x \in {\mathbb Z}(J(G))$.
(5) If ${\cal B} \subseteq {\cal A}(G)$ then
$J( \langle {\cal B} \rangle) =  \langle {\cal B} \rangle $.\\
\begin{quote}
\emph{Proof:}   This is straightforward.
\end{quote}
{\bf Definition 8:}
$G$ is Thompson factorizable
with respect
to $p$ if $G=O_{p'}(G) C_G( \Omega(Z(S))) N_G(J(S))$.
Note that $G$ is Thompson factorizable iff $G/O_{p'}(G)$ is.
\\
\\
{\bf Thompson Factorization:}
Let $O_{p'}(G)=1$ and $V= \langle \Omega({\mathbb Z}(S)): S \in S_p(G) \rangle $.
Then $G$ is Thompson factorizable iff $J(G) \le C_G(V)$.
\begin{quote}
\emph{Proof:}
Let $S \in S_p(G)$ and $C= C_G(V)$.  Assume that $G$ is Thompson factorizable.
$\Omega({\mathbb Z}(S)) \le {\mathbb Z}(J(S))$ and so $V \le {\mathbb Z}(J(S))$ and
$C \le J(G)$.
Assume $C \le J(G)$.  The $J(G) \le C \cap S$.  Since $J(S) \; char \; C \cap S \in S_p(C)$
and
$\Omega({\mathbb Z}(S)) \le {\mathbb Z}(J(S))$, Frattini yields
$G= C N_G(C \cap S) =
C_G( \Omega(Z(S))) N_G(J(S))$.
\end{quote}
{\bf Alternate Thompson subgroup:}
$P$ a $p$-group and set $d(P)= sup\{ |A| : A \le P, A'=1 \}$.  Let
${\cal A}(P)= \{ A: |A|=d(P), A'=1 \}$ then $J(P)= \langle A : A \in {\cal A}(P) \rangle $.
\\
\\
{\bf Lemma:}
(1) $A \in {\cal A}(P) \rightarrow A=C_P(A)$;
(2) $C_P(J(P))= {\mathbb Z}(J(P)) = \bigcap_{A \in {\cal A}(P)} A$;
(3) $H \le P$ and $d(H)=d(P) \rightarrow J(H) \le J(P)$ and
${\mathbb Z}(J(H)) \le {\mathbb Z}(J(P))$;
(4) Suppose $J(P) \subseteq H \subseteq P$ then $J(H)=J(P)$.
\begin{quote}
\emph{Proof:}
If $x$ centralizes $A$, $ \langle x, A \rangle $ is abelian but then
$| \langle x,A \rangle | > |A|$ so $x \in A$; this proves
(1).  $C_P(J(P))= \bigcap_{A \in {\cal A}(P)} C_P(A)$ since $ \langle A \rangle =J(P)$.  Thus
$C_P(J(P)) = \bigcap_{A \in {\cal A}(P)} A \subseteq J(P)$ so
$C_P(J(P))= {\mathbb Z}(J(P))$; this proves (2).  (3) and (4) are clear.
\end{quote}
{\bf Theorem 23:}
Suppose $O_{p'}(G)=1$, $G$ is $p$-stable and $P \in S_p(G)$.  If $H \in {\cal SCN}(P)$ then
$H \subseteq O_p(G)$.
\begin{quote}
\emph{Proof:}
Put $L=O_p(G)$.  Since $L \le P$, $[L, H] \subseteq H$ and so $[L, H, H]=1$ so
$x \in H \rightarrow [L,x,x]=1$ and thus $H/C_G(L) \subseteq O_p(G/C_G(L)$.  Since
$L$ is $p$-constrained, $C_G(L) \subseteq L$ and thus $H \subseteq L$.
\end{quote}
\section{Glauberman's $Z(J)$ Theorem:}
{\bf Glauberman Replacement Lemma:}  If $p \ne 2$, $P$, a $p$-group, $B \lhd P$,
$A \subseteq P$, $A'=1$, $B \nsubseteq N_P(A)$ and $A \cap B \supseteq B'$ then
$\exists A^* \subseteq P, (A^*)'=1$ with
(1) $A^* \cap B > A \cap B$;
(2) $|A^*|= |A|$;
(3) $A^* \subseteq N_P(A) \rightarrow [A^*, A, A]=1$.
\begin{quote}
{\bf Outline of proof:}
(1) Reduce to $P=AB$, $N_P(A) \lhd P$;
(2) $x \in B-N \ne \emptyset, A \cap B \subseteq A \cap A^x$;
(3) $u=A A*$, $V=A \cap A^*, W=U \cap B$;
(4) $[x, A]$ is abelian;
(5) $VW$ is abelian;
(6) $VW$ works.
\\
\\
\emph{Proof:}
Proof is by induction $|P|$.  If $|P|>|AB|$, we are done by induction,
so $P=AB$.  Suppose $N=N_P(A)$, $\exists M$ such that $N \le M \lhd P$ since
the maximal subgroups of $P$ are normal.  Let $B_1= B \cap M$ then by
Dedekind, $A B_1= M$ and $B_i \nsubseteq N_M(A)$.  Applying induction,
we find an $A^*$ such that $A^*$ satisfies (1), (2) and (3) in $A B_1$.  This
$A^*$ works in $P=AB$.  So we may assume, $N \lhd P$.\\
\emph{Claim:} $x \in B-N \rightarrow A \cap B \subseteq A \cap A^x$ and
$A^x \le N$.  Proof of claim:
$A \cap B \supseteq B'$ $\rightarrow A \cap B \lhd B$ $ \rightarrow A \cap B= (A \cap B)^x$
$\rightarrow A \cap B \subseteq A^x \cap B \rightarrow A \cap B \subseteq A \cap A^x$.
Now, $A<N$ so $A^x < N^x=N$ by the above and this proves the claim.
\\
\\
Let $U=A A^x<N$, $V=A \cap A^x$, $W= U \cap B$.  $A^x \subseteq  N_P(A)$ so $U$ is a
group and $Z \lhd U$.\\
Claim:  $U' \subseteq V \subseteq {\mathbb Z}(U)$ and $W= [x,A](A \cap B)$.  Proof of claim:
$A^x \subseteq N, A \subseteq N$ so $A^xA \subseteq N$ and $A^xA, A^xA] \subseteq A \cap A^x$,
which proves the claim.\\
Continuing, $A \cap B \subseteq U \cap B$ and $[x,a]= (a^{-1})^xa \in U$ and
$[x,a]= x^{-1} x^a \in B$ so $[x,A](a \cap B) \subseteq W$.    If $y=a_1 a_2^x \in W$,
$a_1 , a_2 \in A \rightarrow a_1 a_2 [a_2,x]= a_1 a_2^x \in B$ so
$a_1 a_2 \in B$ and $y \in [x,A](A \cap B)$.
\\
\\
\emph{Major claim:} $[x, A]$ is abelian.
\\
Subclaim: $U$ has nilpotence class $\le 2$ ($U' \subseteq {\mathbb Z}(U)$) implies
$[ac,b] [a,b][c,b]$.  This is a calculation.\\
Let $a, a_1 \in A$.  $[x,a,a_1]= [[x, [x,a]^{-1} [x,a], a_1^x]$ but since $[x,a] \in B$,
$[x, [x,a]^{-1}] \in B \subseteq A \cap B \subseteq {\mathbb Z}(U)$,
$[x,a,a_1]^x = [[x,a],a_1^x]$ by the subclaim.
Now, $[[x,a],a_1^x]= [a,a_1^x]= [x,a_1,a]$.
So $\forall a, a_1 \in A: [x,a,a_1]^x=[x,a_1,a] \rightarrow [x,a,a_1]^{x^2}=[x,a,a_1]$
(Equation **).  Since $x$ has odd order, this becomes $[[x,a],a_1^x]= [x,a,a_1][[x,a],[a_1,x]]$.
So $[[x,a],[a_1,x]]=1$; these two generators of $[x,A]$ commute so $[x,A]$ is abelian.
\\
\\
Claim: $VW$ is abelian: $[x,A] \subseteq U$ is abelian and $A \cap B \subseteq {\mathbb Z}(U)$,
so $W= [x,A] (A \cap B)$ is abelian.  Finally, $V \subseteq {\mathbb Z}(U)$, so $VW$ is
abelian.
\\
$A^*=VW$ satisfies the theorem:\\
(1) $[x,A] \nsubseteq A$ since $x \notin N_P(A)$.
$A \cap B < W <B$ so $A^* \cap B > A \cap B$.\\
(2) $|A^*|=
{\frac {|V| |W|} {|V \cap W|}} =
{\frac {|A \cap A^x| |U \cap B|} {|U \cap B \cap A \cap A^x|}} =
{\frac {|A \cap A^x| |U \cap B|} {|A \cap B|}}$.
$|P|=|AB|=
{\frac {|A| |B|} {|A \cap B|}} =
{\frac {|U| |B|} {|U \cap B|}}$.  So
$
{\frac {|U|} {|A|}}=
{\frac {|U \cap B|} {|A \cap B|}}$ and $|A^*|= {\frac {|A \cap A^x| |U|} {|A|}}$.
$|A^*| = {\frac {|A \cap A^x||A A^x|} {|A|}}= {\frac {|A| |A^x|} {|A|}}= |A|$.\\
(3)
$A^* \subseteq U$ since $V, W \subseteq U$ and $U \subseteq N$ since $A, A^x \subseteq N$.
\end{quote}
{\bf Glauberman's $Z(J)$ Theorem:}  Assume $p \ne 2$ is prime and that $G$ is a
$p$-stable, $p$-constrained group with $O_{p'}(G)=1$ and $P \in S_p(G)$ then
${\mathbb Z}(J(P)) \lhd G$.
\begin{quote}
\emph{Proof:}
Let $Z= {\mathbb Z}(J(P))$ and $H= O_p(G)$.  $J(P) \; char ;\ P$ and $Z \; char \; P$ so
$Z \subseteq H$.  Let $H_0$ be a minimal normal $p$-subgroup of $G$ such that $Z \cap H_0$ is not
normal in $G$.  Put $Z_0= H_0 \cap Z$ and let $K/C_G(H_0)= O_p(G/C_G(H_0))$ so $K \lhd G$.
Also put $P_0=P \cap K \in S_p(K)$.
\\
\\
(1) $K= P_0 C_G(H_0)$. Reason: Since $K/C_G(H_0)$ is a $p$-group and $P_0C_G(H_0) \subseteq K$,
$K \ne P_0 C_G(H_0)$ implies that $|K/C_G(H_0)| \ge p |(P_0C_G(H_0))/C_G(H_0)= p {\frac {|P_0 |}
{|P_0 \cap C_G(H_0)|}}$. Thus $|P_0| \mid p \cdot |K|$ which contradicts the fact that
$P_0 \in S_p(K)$.
\\
(2)  By Frattini, $G= KN_G(P_0)$.
\\
(3) $J(P) \nsubseteq P_0$.  Reason: If not, $J(P) < J(P_0) \lhd N_G(P_0)$ so
$Z_0 \lhd G$, a contradicton.
\\
(4) $H_0' \subseteq Z_0$. Proof: Since $H_0$ is a $p$-group $H_0' < H_0$ so by
maximality, $Z \cap H_0' \lhd G$ and $H_0 = \langle Z_0^g \rangle $.
$[Z_0, H_0] \subseteq Z_0 \cap H_0' \lhd G$,
so $[Z_0, H_0]^g = [H_0, H_0] \subseteq Z_0 H_0'$ and $H_0' \subseteq Z_0 \cap H_0'$.
\\
(5) $H_0 \nsubseteq N_G(A)$ otherwise, $[H_0,A] \subseteq A$ and
$[H,A,A]=1$ which implies $A \subseteq K$ by $p$-stability.
\\
(6)  $H_0 \subseteq {\mathbb Z}(J(P_0))$.
\\
\\
Conclusion:
By (1) and (2), $G= N_G(P_0) C_G(H_0)$ so $Z_0^g=Z_0^n, n \in N_G(P_0)$ and
$H_0= \langle Z_0^g \rangle \subseteq {\mathbb Z}(J(P_0)), \forall g \in G$.  Therefore
$H_0 \subseteq A^*$ and $A^* \subseteq N_G(A)$ so $H_0 \subseteq N_G(A)$ which is
a contradiction.
\end{quote}
\section{Alperin-Lyons Proof of Baer}
{\bf Theorem 24:}  Suppose $x$ is a $p$-element of $G$.
$x \in O_p(G)$ iff $ \langle x, x^g \rangle $ is a $\forall g \in G$.
\begin{quote}
\emph{Proof:}
Let $K= \langle x^G \rangle $ and suppose $z, y \in K \rightarrow \langle z, y \rangle $
is a $p$-group.  Suppose $K \nsubseteq O_p(G)$.
\\
\\
(1) $\exists P, Q \in S_p(G): P \cap K \ne Q \cap K$.
Reason: If $K \subset R, \forall R \in S_p(G)$ then $K \subseteq  \bigcap R^g \lhd G$ so
$K \subseteq O_p(G)$.
\\
\\
(2) $|P \cap K| = |Q \cap K|$.  Reason: $Q= P^g$ so $Q \cap K= P^g \cap K^g= (P \cap K)^g$
so $|Q \cap K|= |P \cap K|$.
\\
\\
(3) $K \cap P \nsubseteq Q$ and $K \cap Q \nsubseteq P$ for $P, Q$ chosen such that
$P \cap K \ne Q \cap K$.  If $K \cap P \subseteq Q$, $K \cap P \subseteq K \cap Q$ if
since the cardinalities are the same, $K \cap P = K \cap Q$.
\\
\\
(4) We can choose $P$ and $Q$ such that $|K \cap P \cap Q|$ is maximal with respect to
$K \cap P \ne K \cap Q$.  Put $W=P_0< P_1< \ldots < P_n=p$ with $|P_i:P_{i-1}|=p$.
Now $P \cap K \nsubseteq W$ (otherwise $ \langle P \cap K \rangle \subset P \cap Q$ so
$P \cap K \subset Q$
which is a contradiction).
Let $j$ be the smallest $j$ such that $P_j \cap K \nsubseteq W \cap K$ ($ j \ge 1$).  Pick
$x \in K\cap P_j-W$.
\\
\\
(5) Claim: $x \in N(W)$.  Proof of claim: $x$ normalizes $P_{j-1}$ so $P_{j-1} \cap K$
is normalized by $x$.  By the choice of $j$, $P_{j-1} \cap K \subseteq W \cap K$ so
$P_{j-1} \cap K = W \cap K$.
Thus $W= \langle P_{j-1} \cap K \rangle = P_{j-1} \cap K \rangle^x= W^x$ which
proves the claim.\\
For the same reason, $\exists y \notin W: y \in K, y \in N(W)$ and
$ \langle x, y \rangle $ is a $p$-group.
Thus if $R \in S_p(G)$ with $ \langle x, y \rangle W  \subset R$,
$K \cap P \cap R \supseteq K \cap W \cap \langle x \rangle $ and
$|K \cap P \cap R| > |K \cap P \cap Q|$
and symmetrically,
$|K \cap Q \cap R| > |K \cap P \cap Q|$ thus $P \cap K = R \cap K= K \cap Q$ concluding the proof.
\end{quote}
\section{Goldschmidt's Proof of Burnside's Theorem for $p \ne 2$}
{\bf Burnside's Theorem:}  If $|G|= p^a q^b$ for $p \ne 2 \ne q$ then $G$ is solvable.
\begin{quote}
\emph{Proof:}
Let $G$ be a minimal counterexample, $r \in \{ p , q \}$.
\\
\\
\emph {Lemma 1:}  If $R \in S_r(G)$ then if $1 \ne S \in r'(G)$, $R \nsubseteq N_G(S)$.\\
Proof: Let $Q \in S_{r'}(G)$ with $S \subseteq Q$.  Since $G=RQ$, $\forall g, Q^g=Q^r$ for some
$r \in R$.  Now suppose $R \subseteq N_G(S)$ then $S \subseteq Q^r$ and thus
$1 \ne \bigcap_{r \in R} Q^r = \bigcap_{g \in G} Q^g \lhd G$ which is impossible since $G$
is simple.
\\
\\
\emph {Lemma 2:}  If $M$ is a maximal subgroup of $G$ then $F(M)$ is an $r$-group.\\
Proof:
From now on, let $F=F(M)= F_p \times F_q$, and $Z={\mathbb Z}(F) = Z_p \times Z_q$.
Observe that if $M$ is maximal, $M$ is solvable and so
$O_{p'}(N_M(P)) \subseteq O_{p'}(M)$.
\\
\\
\emph{Claim 1:}  $F$ is not cyclic.\\
Assume, by way of contradiction, that $F$ is cyclic.
Suppose $q > p$ and $Q \in S_q(M)$.  Since $Q$ acts on $F_p$ and $q \nmid Aut(F_p)$, $[Q, F_p]=1$.
$Q$ also acts on $F_q$ and
$Q/C_Q(F) \rightarrow Aut(F_q)$.
$C_M(F) \subseteq F$, since $M$ is solvable so $C_M(F) \subseteq F$ and $C_Q(F) \subseteq F_q$.  Clearly,
$F_q \subseteq C_Q(F)$ since $F_q$ is cyclic. So $Q/F_q \subseteq Aut(F_q)$.  Further, $F_q \thinspace char \thinspace Q$ and
$N_G(Q) \subseteq N_G(F_q)$.  $N_G(Q) \subseteq N_G(F_q) = M $.  Examining $N_M(F_q)/C_M(F_q) \subseteq Aut(F_q)$, we see $|N_M(F_q)|_q = |Q|$ so
$|N_M(F_q)|_q = |N_G(F_q)|_q$.
Thus, $Q \in S_q(G)$, contradicting Lemma 1.
\\
\\
\emph{Claim 2:}  $M$ is the unique maximal subgroup containing $Z$.
\\
Suppose $Z \subseteq M_1 \ne M$. $M_1$ a maximal subgroup of $G$.
$M= N_G(Z_p)= N_G(Z_q)$, $Z_p \subseteq O_{q'}(N_{M_1}(Z_q))$ and
$Z_q \subseteq O_{p'}(N_{M_1}(Z_p))$.
Because $M$, is solvable, $O_{p'}(N_M(F)) \subseteq O_{p'}(M)$,
$Z_p \subseteq F(M_1)_p \subseteq C(Z_q) \subseteq M$ and
$Z_q \subseteq F(M_1)_q \subseteq C(Z_p) \subseteq M$ so
$F(M_1) \subseteq F(M)$.  Since
the argument also applies to $M_1$, $F(M) \subseteq F(M_1)$. So
$F(M)= F(M_1)$.  This $M= N_G(F(M)) = M_1$, proving the claim.
\\
\\
Now, since $F$ is not cyclic, there is an Abelian subgroup
$V \subseteq F$ of type $(r,r)$.  $\forall x \in V^{\#}$, $Z \subseteq C(x)$
and by the uniqueness of $M$, $C(x) \subseteq M$.  Let $V \subseteq R \in S_r(M)$.
If $Q_0 \in r'(G)$  with $R \subseteq N(Q_0)$.  $Q_0 = \prod_{x \in V^{\#}} C_{Q_0}(x) \subseteq M$.
It follows that $F_{r'}$ is the unique maximal $r'$-subgroup of $G$ normalized by $R$, so
$N(R) \subseteq N(F_{r'}) = M$ so $R \in S_r(G)$, contradicting Lemma 1.  This proves Lemma 2.
\\
\\
\emph {Lemma 3:}  If $R \in S_r(G)$ then $R$ is contained in a unique maximal subgroup of
$G$.  Every maximal subgroup, $M \subseteq G$, contains a Sylow subgroup of $G$.
\\
\emph{Proof:} $R \subseteq M$ so $O_{p'}(M)=1$ by lemma 1.  Since $M$ is
$r$-constrained, $O_{r'}(M)=1$ and $M$ is solvable, $M$ is $r$-constrained.  Once
we know $M$ is $r$-stable, by Glauberman's $Z(J)$ theorem, we have: $M=N_G({\mathbb Z}(J(R)))$.
$p^aq^b$ is odd and $|SL_2(r)|= (r^2-1)r$ which is even, so $G$ is $p$-stable and $M=N_G({\mathbb Z}(J(R)))$.
So $M$ is unique.
Let $M_1$ be any maximal subgroup of $G$, we can choose $r$, such that $O_{r'}(M_1)=1$.
If $R_1 \in S_r(M)$, we have
$M_1 =N_G({\mathbb Z}(J(R_1)))$.
Since ${\mathbb Z}(J(R_1)) \; char \; R_1$,
$N_G(R_1) \subseteq M_1$, so $R_1 \in S_r(G)$ ,proving the second statement.
\\
\\
\emph {Lemma 4:}  If $R \in S_r(G)$ then ${\mathbb Z}(R)$ is contained in a
unique maximal subgroup of $G$.
\\
\emph{Proof:}  Suppose ${\mathbb Z}(R) \subseteq M \cap M_1, M \ne M_1$ with $M, M_1$ maximal in $G$.
We may assume $M_1$ is chosen such that $|M \cap M_1|_r$ is maximal.  Now, let
$R_1 \in S_r(M \cap M_1)$ such that ${\mathbb Z}(R) \subseteq R_1$.  By the maximality of $|M \cap M_1|_r$,
$N_G(R_1) \subseteq M$ and $R_1 \in S_r(M_1)$
Conjugating $R$ by something in $M$, we may assume, by lemma 1,
$R_1 \subset R$ and $M_1$ contains a $r'$ sylow subgroup of $G$.
Thus $G=RM_1$ and
$1 \ne {\mathbb Z}(R) \subseteq \bigcap_{r \in R} M_1^r= \bigcap_{g \in G} M_1^g \lhd G$,
a contradiction.
\\
\\
\emph {Lemma 5:}  $\exists R_1, R_2 \in S_r(G)$ such that $R_1 \cap R_2 = 1$.
\\
\emph{Proof:}  Let $R_1 \in S_r(G)$ and let $M$ be a unique maximal subgroup containing
$R_1$.  Pick $R_2 \nsubseteq M$.  We can do this since $G$ is simple.\\
\\
Claim: $R_2 \cap M=1$.
\\
If not, choose $R_2$ such that $|R_2 \cap M|$ is as large as possible.  Put
$R_0 = (R_2 \cap M)$.
We can assume, possibly after conjugation in $M$, $R_0 \subseteq R_1$.  ${\mathbb Z}(R_1) \subseteq N(R_0)$.
By lemma 4, $N(R_0) \subseteq M$.  $R_0 < R_2$ so $N_{R_2}(R_0) > R_0$ and $|R_2 \cap M| > |R_0|$,
which is a contradiction.  So $R_2 \cap M=1$.
\\
\\
Conclusion: WLOG, assume $q^b > p^a$ and let $Q_1, Q_2 \in S_q(G)$ with $Q_1 \cap Q_2=1$.
Then $|G| \ge |Q_1| \cdot |Q_2| > |G|$ which is just plain wrong.
\end{quote}
\section{Thompson Complements}
{\bf Thompson Normal $p-$Complement:}
Let $p \ne 2$ and $P \in S_p(G)$.  Assume $N_G(J(P))$
and $C_G(\Omega_1({\mathbb Z}(P)))$ have a normal $p-$complement then so does $G$.
\begin{quote}
\emph{Proof:}
Let $G$ be a counterexample of minimum order.
\\
\\
\emph{Step 1:}  Let ${\cal H}= \{ H \in p(G): N(H)$ does not have a normal $p$-complement $\}$.
If $H, K \in {\cal H}$.  We say $H \le K$ if one of the following holds:
(1) $|N(H)|_p < |N(K)|_p$;
(2) $|N(H)|_p = |N(K)|_p$ and $|H| < |K|$; or,
(3) $|H| = |K|$.  Choose $H$ minimal with respect to the ordering and set
$N=N(H)$.  Let $Q \in S_p(N)$ and $H \subseteq Q \subseteq P$.
\\
\\
\emph{Step 2:} $H \ne P$
\\
If $H=P$, $N \subseteq N(J(P))$ which is a contradiction.
\\
\\
\emph{Step 3:} ${\overline N}= N/H$ has a normal $p$-complement.
\\
Let ${\overline Q}= P/H$ and ${\overline Q} \in S_p({\overline N})$.  Suppose
${\overline N}$ does not have a
normal $p$-complement.
Since $|{\overline N}| < |{\overline G}|$, either
$C_{\overline N}({\mathbb Z}({\overline P}))$ or
$N_{\overline N}(J({\overline Q}))$ does not have a
normal $p$-complement.  Let $K$ be the inverse image of the one that fails to have a
normal $p$-complement.  $K \in p(G)$ and $N(K)$ has no
normal $p$-complement.  Since $Q \subseteq N(K)$, either
$|N(K)|_p > |N(H)|_p$ or
$|N(K)|_p = |N(H)|_p$ and $|H| < |K|$.  Hence, $K \ge \ge K$.  But $K \ne H$ and this
contradicts maximality of $H$ in ${\cal H}$.
\\
\\
\emph{Step 4:} $N=G$.
\\
$N$ satisfies the hypothesis of the theorem $H \subseteq Q \subseteq P$ so
${\mathbb Z}(P) \subseteq  N(H) =N$.
$Q {\mathbb Z}(P) \subseteq N \cap P$ so $Q {\mathbb Z}(Q) = Q$ and
${\mathbb Z}(P) \subseteq Q$ thus $
{\mathbb Z}(P) \subseteq
{\mathbb Z}(Q)$ and $C_N({\mathbb Z}(Q)) \subseteq
C_G({\mathbb Z}(Q)) \subseteq C_G({\mathbb Z}(P))$.
If $P= Q$, $N(J(Q)) \subseteq N(J(P))$ has a
normal $p$-complement.
Suppose $Q < P$  so $N_P(Q) > Q$ and $|N_G(J(Q))|_p > |N|_p = |N_G(H)|_p$.
By the maximality of $H$ in ${\cal H}$, $J(Q) \notin {\cal H}$ and so
$N(J(Q))$ has a
normal $p$-complement and
so does $N_N(J(Q))$.  If $ N \ne G$, then by the minimality of $|G|$, $N$ has a
normal $p$-complement and so $N=G$.
\\
\\
\emph{Step 5:}  $O_{p'}(G)=1$.
\\
Let $J= O_{p'}(G)$ and ${\overline X} = X/L$.  If $K \subset P$,
$ {\overline {N(K)}} = N({\overline K})$,
${\overline {J(P)}} = J({\overline P})$,
${\overline {C(P)}} = C({\overline P})$, and
${\overline {{\mathbb Z}(P)}} = {\mathbb Z}({\overline P})$.  So
$N({\mathbb Z}(P))$ has a
normal $p$-complement.  If $|{\overline G}| < |G|$, ${\overline G}$ and the inverse image is
a normal $p$-complement of $G$.
\\
\\
\emph{Step 6:}  $H= O_p(G)$ and $G$ is $p$-solvable of $p$-length at most $2$.
\\
Set $K= O_p(G), K \subseteq H$.  Since $G$ has no
a normal $p$-complement,  $N(K)=G$ and $K \in {\cal H}$, by steps 3, 4,
$G/H$ has a
normal $p$-complement so $G$ has $p$-length $\le 2$.
\\
\\
\emph{Step 7:}  If ${\overline G}= G/H= {\overline P}{\overline M}$ where ${\overline M}$
is a normal $p$-complement
and ${\overline M}$ contains no $P$-invariant subgroup.
\\
Suppose ${\overline M} > {\overline M}_0 >1$ and
${\overline M}_0$ is $P$-invariant.  Let $M_0$ be the inverse image and set
$G_0= P M_0$.  $G_0 < G$ so $G_0$ has a
normal $p$-complement
and $K_0 \lhd G_0, K_0 \cap P = 1$.  $[H, K_0] \subseteq K_0 \cap H =1$ so
$K_0 \subseteq C(H)$.  But then $K_0 \subseteq {\mathbb Z}(O_p(G))$ by Hall-Higman which is
a contradiction.
\\
\\
\emph{Step 8:} ${\overline M} = {\overline R}$ is an elementary abelian $r$-group for some
$r \ne p$ and $P$ acts irreducibly on ${\overline R}$. Hence, $P$ is maximal in $G$.
\\
Let $r \mid |{\overline M}|$.  $P$ permutes Sylow $r$-subgroups and by Sylow, the conjugacy
class has size $1$.   So ${\overline R} \in S_r({\overline M})$ is $P$-invariant and since
${\overline R}$ has no non-trivial characteristic subgroups, it is elementary abelian.
Since ${\overline M}$ has no proper ${\overline P}$ invariant subgroups, ${\overline P}$
acts irreducibly.
If $G > L >P$, $1 \le {\overline L} \cap {\overline R} < {\overline R}$
and ${\overline L} < {\overline R}$ is a proper ${\overline P}$-invariant subgroup of
${\overline R}$ which is a contradiction.
\\
\\
\emph{Step 9:} $\exists 1 \ne A \in P$ with $A$ abelian such that $m_p(A)= d(P)$ and
$A \nsubseteq H$.
\\
Let $A$ be a fixed one of minimal order.  If $A_0= A \cap H$,
$A/A_0$ is elementary abelian.  If $J(P) \subseteq H$ then $J(P) \; char \; H$ and
$J(P) \lhd G$ which is a contradiction.  So $J(P) \nsubseteq H$ and such an $A$ exists.
Choose $A$ as mentioned and set $A_0= A \cap H$,  $A_1= \Omega_1(A/A_0)$, $A_1 \nsubseteq H$
and $m(A)= m(A_1)$ so $A= A_1$ and $A/A_0$ is elementary abelian.
\\
\\
\emph{Step 10:}  Let ${\overline A}= (AH)/H$ then $
{\overline G} = {\overline A} {\overline Q} $ and
$|{\overline A} |=p$.
\\
${\overline G}= {\overline P} {\overline R}$ and
${\overline P}$ normalizes
${\overline R}$.  The action is faithful since $H=O_p(G)$ and the kernel would be a normal
$p$-subgroup of $G$.  Note ${\overline A}= (AH)/H \cong A/A_0 \ne 1$.
${\overline A}$ acts nontrivially on
${\overline R}$ by the previous result an we can find
$ {\overline R}_1 \subseteq {\overline R}$ on which ${\overline A}$ acts non-trivially and
irreducibly.  Let $G_1$ be the inverse image of
$ {\overline A} {\overline R} $ and $P_1 \in S_p(G_1)$.   $A \subseteq P_1 \subseteq P$.
Since $H \subseteq P_1$ and
$C_G(H) \subseteq H$,
${\mathbb Z}(P) \subseteq C_G(H) \subseteq H \subseteq P$, and
${\mathbb Z}(P) \subseteq {\mathbb Z}(P_1)$ and the latter has a
normal $p$-complement.  Since
$A \subseteq P_1$ and $m(A)= d(P)$, $d(P_1)=  d(P)$ and
$A \subseteq J(P_1 )$.  Let $R_1 \in S_r(N_G(J(P_1 )))$, then
$[A, R_1 ]= [J(P_1 ), Q_2] \subseteq J(P_1 )$ so
$[A, R_1]$ is a $p$ group.  Since
$ {\overline R}_2 \subseteq {\overline R}_1 $,
$ [A, {\overline R}_2] \subseteq {\overline R}_1 $ is an $r$-group
and so
$ [A, {\overline R}_2] =1$.   Thus, $R_2$ is an ${\overline A}$-invariant
subgroup of ${\overline R}_1$ centralized by
${\overline A}$ and ${\overline R}_2 =1$ and $R_2=1$.  So $N_{G_1}(J(P_1 ))$ is a
$p$-group and has a
normal $p$-complement. If $G_1 < G$ then $G_1$ has a
normal $p$-complement and would centralize $H$.   Contradiction.  Hence,
$G = G_1$ and
$ {\overline G} = {\overline A} {\overline R} $
and
$ {\overline A} $ acts faithfully and irreducibly on
${\overline R}$.
$ {\overline A} $ is elementary abelian and hence cyclic so $|A|= p$.
\\
\\
\emph{Step 11:}  Set $W= {\mathbb Z}(H), Z= \Omega_1(W)$.  If $R \in S_r(G)$ then
$[R, Z] \subseteq Z$ but $[R, Z] \ne 1$.
\\
${\mathbb Z}(P) \subseteq C_G(H) \subseteq H$ so ${\mathbb Z}(P) \subseteq W$.
If $[R, W] = 1$, $[R, {\mathbb Z}(P)] = 1$, hence
${\mathbb Z}(P)$ would be central in $G$ which is impossible.
\\
\\
\emph{Step 12:} Contradiction
\\
$Z \lhd G$ and $G$ acts by conjugation so the kernel of the action is $C(Z) \supset H$
and so ${\overline G}$ acts on $Z$.
Since ${\overline R}$ is the unique minimal normal subgroup of ${\overline G}$ and
${\overline G}$ acts nontrivially,
${\overline G}$ acts irreducibly on $Z$ by step 11.
By a previous result, $Z= C_Z({\overline R}) + V$ where $V$ is
${\overline R}$-invariant.
Moreover,
$ {\overline R} \lhd {\overline G} $ and so $V$ is
$ {\overline G} $ invariant and
$ {\overline G} $ acts faithfully on $V$.
Since $|A|=p$, $m(A_0) \ge d(P) -1$.  Set $V_0= V \cap A_0$ and let $t= m(V_0 )$ and
$r= m(A/A_0)$.  $V \subseteq {\mathbb Z}(H)$ so $ \langle V, A_0 \rangle $ is abelian.
Since $V$ is elementary, $d(P) \ge m( \langle V, A_0 \rangle) = m(V)+m(A_0)- m(V \cap A_0)= t+r +
m(A_0)-t= d(P)-1+r$.  Hence, $r=0$ or $r=1$ and $V_0=V$ or $V_0$ is maximal.
Choose $a \in A \setminus A_0$, ${\overline b} \in {\overline R}^H$
such that
$[a, {\overline b}] \ne 1$,
$[{\overline a}, V_0] = 1$ and
$[{\overline a}^{\overline b}, V_0^{\overline b}] = 1$ then
$[\langle {\overline a}, {\overline a}^{\overline b} \rangle , V_0 \cap V_0^{\overline b}] = 1$
and $V_0 \cap V_0^{\overline b} \le p^2$.  Since $A= \langle a \rangle $ is maximal in $Z$ and
${\overline a}^{\overline b} \notin {\overline A}$,
${\overline G}_1, V_0 \cap V_0^{\overline b} = 1$ hence $|V| \le p^2$.
So ${\overline G} \subseteq GL_2(p)$ and since
${\overline G}$ is generated by $2$ elements of order $p$, ${\overline G} \subseteq SL_2(p)$.
So we have an abelian $p'$ group, ${\overline R}$, which is normalized but not centralized by
${\overline A} \subseteq SL_2(p)$ of order $p$ and this is impossible.
\end{quote}
{\bf Theorem (Frobenius):}  If $G$ is solvable and $|G| > 1$,
$\exists p, P \in p(G): 1 \ne P \lhd G$.
\begin{quote}
\emph{Proof:}
The proof is by induction.  It is clearly true for all $p$-groups.   Since $G$ is solvable,
$exists N \lhd G$ and we can assume $|G:N|= p \mid |G|$.  By induction, $exists q:  Q \in q(N), Q \lhd N$.
So $1 \ne O_q(N) \thinspace char \thinspace N \lhd G$.
\end{quote}
{\bf Theorem 25:}
If $P \in S_p(G)$, $G$ contains a normal $p$-complement iff
whenever two elements in $P$ are $G$-conjugate, they are $P$ conjugate.
\begin{quote}
\emph{Proof:}
See the section on transfer.
\end{quote}
{\bf Theorem 26:}
If $G$ has a maximal subgroup, $M$, which is nilpotent of odd order, then $G$ is solvable.
\begin{quote}
\emph{Proof:}
See Passman.
\end{quote}
{\bf Theorem 27:}
Let $p \ne 2$, $P \in S_p(G)$ and suppose for any $H<G: H \; char ;\ P$, $N_G(H)/C_G(H)$ is
a $p$-group, then $G$ has a normal $p$ complement.
\begin{quote}
\emph{Proof:}
By induction on $|G|$.  $N_G(H)$ has a normal $p$-complement by induction.
If $a \in p'(H)$ then $[a,H]=1$.  So $H$ has  normal $p$-complement.
The result follows from Thompson's
normal $p$-complement theorem.
\end{quote}
{\bf Theorem 28:}
Let $p \ne 2$, $G= SL_2(p)$.  The only abelian $p'$-subgroups of $G$ which are normalized
by an $S_p$ subgroup of $G$ lie in ${\mathbb Z}(G)$.
\begin{quote}
\emph{Proof:}
See earlier.
\end{quote}


\section{Terms and Definitions}
\label{sec:terms_and_definitions}

  In this section, we will give a brief overview of the terms and definitions
  that we will use throughout this report. These definitions should not be
  considered complete or even absolutely accurate, as they are a personal
  choice of words with the goal of giving the reader an impression of what
  we are talking about.

  \subsection{Neural networks}
  \label{sub:neural_networks}

    A \textit{Neuron} is a programmatic structure (or function) that is based
    on neurons inside the brain. In its most basic form, it takes a vector of
    input variables, calculates a weighted sum and applies an activation function
    (typically a sigmoid function like \textit{tanh}).

    A \textit{neural network} is a collection of neurons which are typically
    arranged in layers, where the input of a layer is the output of the
    preceeding one. The output of the last layer is called the output of
    the network.

  \subsection{Recurrent Neural Networks}
  \label{sub:recurrent_neural_networks}

    A \textit{recurrent neural network} is a neural network where the
    output of one or more layers is fed back as input to the network
    or a selection of units. This allows the network to make decisions
    based on previous results / events.

  \subsection{Embeddings}
  \label{sub:embeddings}

    TODO

  \subsection{Keras}
  \label{sub:keras}

    Keras is a Python module that contains functions and classes to easily create
    and manipulate neural networks. It contains not only basic neurons, but
    also more involved units like LSTM-Cells or Embedding-Layers (plus
    many more).


\section{Security of the protocol}\label{sec:protosecurity}

Signed and GCM encrypted messages require some attentions in order to avoid
security flaws.

GCM needs to be initialized with a key and an initialization vector (IV).
Moreover, to be secure, the same IV must not be used more than once with the
same key.

The application we developed, derives the GCM key and the first IV using the
secret shared with the algorithm described in \secref{sec:keyxchng}. The GCM key
remains the same for the entire session, while the IV changes after every
encryption/decryption of a message.

The GCM key and the first IV are derived from the secret in this way: first, the
shared secret is hashed using SHA-256; Then, the first 16 bytes of the hash are
used as the GCM key while the last 12 bytes are used as the first IV (4 bytes
are discarded).

Then, after each message (either sent or received), a new IV is computed by
doing the SHA-256 hash of a 20 bytes array constructed by concatenating:
\begin{enumerate*}[label=\textnormal{\arabic*)}]
	\item an incremental counter;
	\item the previous IV\@;
	\item the \code{nonce} field of the header of the last message received
		or sent.
\end{enumerate*}
From the resulting hash the 12 bytes in the center (leaving 10 bytes to the left
and 10 bytes to the right) are taken as the new IV\@.

In this way the next IV can be deterministically derived from the previous IV\@.

Regarding signed message, we need to adopt some additional protection against
replay attacks. An attacker may record the messages exchanged between two
parties \exgratia{the client and the server} and resend some of the recorded
messages to one party in order to impersonate the other. In this case the
receiver will consider the message valid, since it comes with a valid signature.

As said before, the \code{nonce} field in the header ensures randomness in every
message, but this is not sufficient: the nonce is not checked against a list of
previously received nonces, so a replay attack will not be recognized (the
\code{counter} field is used to avoid replay attacks only in a single session:
it cannot prevent replays between two different sessions since when the
application is restarted the counter is initialized back to \(1\)).

To solve this issue, we have introduced the \code{prev\_hash} header field. This
field must always contains the hash of the last received message. The sender of
a message is sure about the freshness of the message he sent thanks to the
\code{nonce} field. So, when he receives the next message, he is sure about the
freshness of the \code{prev\_hash} field. This guarantees that the signed
message is fresh.

Since those fields are also included in GCM encrypted messages, they also ensure
the freshness of an encrypted message.

A formal description of the protocol, using BAN logic, is provided in the next
section.


%%!TEX TS-program = latex
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\usepackage{etex}
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\begin{document}
\onehalfspace

\vspace*{\fill}
\begingroup
\centering

\Large {\scshape Introduction to Statistics}\\

(Lectures 9-10: Testing)

\endgroup
\vspace*{\fill}

\newpage

\section{Testing}

{\scshape Overview:} \noindent Lectures 9-10 will focus on \emph{testing}. The parameter space in the testing game is partitioned into two subsets: the ``null’’ hypothesis $\Theta_0$ and the ``alternative’’ hypothesis $\Theta_1$. There are two actions $\{0,1\}$, where action $a=1$ is interpreted as ``reject the null in favor of the alternative’’  and action $a=0$ is its negation. The loss function used for this problem is the ``0-1 loss’’: $\mathcal{L}(0,\theta) = 0 =  L(1,\theta’)$ for any $\theta \in \Theta_0$, $\theta’ \in \Theta_1$ and $\mathcal{L}(a,\theta)=1$ otherwise. This loss gives rise to the popular \emph{Type I/Type II} error performance criterion.

We first study an idealized environment in which the parameter space has only two components: the null $(\theta_0)$ and the alternative ($\theta_1$). We show that in this set-up any reasonable test must maximize “power” subject to a “size” constraint. Moreover, we show that power maximizer tests reject whenever the “likelihood ratio” between the alternative and the null is large enough.

We then analyze some commonly used testing strategies for parametric models with “composite” null or alternative hypothesis: the Likelihood Ratio test, the Wald test, and the Score test. We specialize these trinity of tests to the linear regression model with unknown variance.

\subsection{The Testing Problem}

Let $X$ be a random variable and let $\{f(x | \theta)\}_{\theta \in \Theta}$ be a statistical model.\footnote{Throughout this section, we will work with statistical models in which $f(x | \theta)$ is a p.d.f.} Partition the parameter space into two subsets $\Theta_0$ and $\Theta_1$.\footnote{By partition, we mean $\Theta_0 \cap \Theta_1 = \emptyset$ and $\Theta_0 \cup \Theta_1 = \Theta$.}The testing problem starts as follows. In the first stage nature selects a parameter $\theta \in \Theta$. Since the null and the alternative hypothesis partition the parameter space, then either $\theta \in \Theta_0$ or $\theta \in \Theta_1$, but not both.

As usual, the econometrician cannot observe the parameter selected by nature but observes data. Based on the data, the statistician would like to decide whether the parameter selected by nature belongs to $\Theta_0$ (the null) or to $\Theta_1$ (the alternative).

In the simplest set-up, the statistician can take only two actions: $a \in \{0,1\}$. If the econometrician picks $a=1$, he/she will be rejecting the null in favor of the alternative (and thus, saying that $\theta \in \Theta_1$). If, however, $a=0$ we the econometrician does not reject the null (which we interpret as saying that $\theta \in \Theta_0$).

A \emph{decision rule/algorithm/strategy} for the statistician in the hypothesis testing problem is a mapping
\begin{equation*}
\phi:X \rightarrow \{0,1\}
\end{equation*}
These decision rules are called ``tests’’. The collection of all data sets for which the econometrician rejects the null hypothesis
$$\{x \: | \: \phi(x)=1\} $$
\noindent is called the critical region of the test $\phi$. \\

\subsection{0-1 Loss and Type I/Type II error}

The payoff/loss for the econometrician depends on the action taken (1 or 0) and the true state of the world ($\theta$). The following $\{0,1\}$-loss structure is typical in testing problems:
\begin{equation*}
\left.\begin{array}{ccc} a/s& \theta \in \Theta_0 & \theta \in \Theta_1 \\ 1 & 1 & 0 \\0 & 0 & 1\end{array}\right.
\end{equation*}

This is intended to model a payoff structure in which the econometrician is only punished when taking the ``wrong’’ decision. The punishments are symmetric, but it is easy to extend this framework.

The expected loss (risk) is computed as follows. When $\theta \in \Theta_0$ the econometrician only experiences a loss if he/she (incorrectly) rejects the null hypothesis. Thus, the expected loss equals the probability of (incorrectly!) rejecting the null hypothesis ($\phi(x)=1$) when the null hypothesis is true. For any $\theta \in \Theta_0:$

\begin{equation} \label{equation:TypeI}
\mathbb{E}_{f(x|\theta)} [  \mathcal{L}(\phi(x),\theta)   ] = \mathbb{E}_{f(x|\theta)} [ \mathbf{1}\{\phi(x)=1\}   ] = P_{\theta}(\phi(x)=1) .
\end{equation}

\noindent This is referred to as the \emph{rate of Type I error} of the test $\phi(x)$ at $\theta \in \Theta_0$. The \emph{largest} rate of Type I error of a test; i.e.,

\[ \sup_{\theta \in \Theta_0} P_{\theta}(\phi(x)=1)  \]

\noindent is called the \emph{size} of the test.

When $\theta \in \Theta_1$ the econometrician only experiences a loss if he/she (incorrectly) fails to reject the null hypothesis. Thus, the expected loss equals the probability of (incorrectly!) failing to reject the null hypothesis ($\phi(x)=0$) when the null hypothesis is not true. For any $\theta \in \Theta_1$:
\begin{equation} \label{equation:TypeII}
\mathbb{E}_{f(x|\theta)} [  \mathcal{L}(\phi(x),\theta)   ] = \mathbb{E}_{f(x|\theta)} [ \mathbf{1}\{\phi(x)=0\}   ] = P_{\theta}(\phi(x)=0) .
\end{equation}

\noindent This is the \emph{rate of Type II error} of the test $\phi(x)$ at $\theta$. It is also common to make reference to the \emph{power} of a test at $\theta \in \Theta_1$. The power is defined as the probability of rejecting the null when it is not true; thus, it equals one minus the rate of Type II error; i.e.,

\[ 1- P_{\theta} (\phi(x)=0) = P_{\theta}(\phi(x)=1). \]

\subsection{Testing a ``simple’’ null against a ``simple’’ alternative}

Suppose that both $\Theta_0$ and $\Theta_1$ are singletons, so that the testing problem becomes
\[ \textbf{H}_0: \theta=\theta_0 \textrm{ vs. }  \textbf{H}_1: \theta=\theta_1. \]
In this problem it is relatively straightforward to characterize the ``optimal’’ test. Define a randomized test as a map
$$\phi:X \rightarrow [0,1]$$
\noindent where $\phi(x)$ is interpreted as the probability of ``rejecting the null hypothesis’’ after observing $x$. See \cite{Ferguson67}, p. 198, 199 (also, think about what we did in the last problem of problem set 3!). The rates of Type I/Type II error of a randomized test $\phi$ are given by:
\[ R(\phi,\theta_0) = \mathbb{E}_{f(x|\theta_0)}[\phi(x)], \quad R(\phi,\theta_1) = 1-\mathbb{E}_{f(x|\theta_1)}[\phi(x)]. \]

The following proposition shows that under a mild assumption, any admissible test for this problem maximizes power subject to a ``size control’’ constraint.

\begin{proposition} Suppose that for any set $A \subseteq \mathbf{X}$
$$\int_{A} f(x,\theta_0)dx > 0 \implies \int_{A} f(x,\theta_1)dx > 0 .$$
A randomized test $\phi$ is admissible if and only if there exists $\alpha \in [0,1]$ such that $\phi$ maximizes power subject to having size at most $\alpha$; that is
\begin{equation}
\phi \in \arg \max_{\phi} \left( 1-R(\phi, \theta_1) \right) \label{equation:optimization}
\end{equation}
\noindent s.t.
\begin{equation}
R(\phi, \theta_0) \leq \alpha \label{equation:sizecontrol}
\end{equation}
\end{proposition}
\begin{proof}
\noindent ``$\Rightarrow$'' First we show that if $\phi$ is admissible, then $\phi$ solves $(\ref{equation:optimization})$ subject to  (\ref{equation:sizecontrol}) for some $\alpha$. We show that contrapositive. Let $\alpha$ be the rate of Type I error of $\phi$. Suppose $\phi$ does not solve the constrained optimization problem for any such value of $\alpha$. Then, there is another test $\phi'$ (namely, the solution to the constrained optimization problem for $\alpha = R(\phi,\theta_0)$) such that  $R(\phi,\theta_1) > R(\phi',\theta_1) $ and $R(\phi,\theta_0) \geq R(\phi',\theta_0)$. Hence, $\phi$ is not admissible. \\

\noindent ``$\Leftarrow$'' We do the proof by contradiction. Suppose that $\phi$ solves $(\ref{equation:optimization})$ subject to  (\ref{equation:sizecontrol}) for some $\alpha$, but is not admissible. Then, there exists $\phi'$ such that $R(\phi,\theta_1) \geq R(\phi',\theta_1) $ and $R(\phi,\theta_0) \geq R(\phi',\theta_0)$ with strict inequality somewhere. We have three cases to consider:
\begin{enumerate}
\item If $R(\phi,\theta_0)=R(\phi', \theta_0)$ and $R(\phi,\theta_1) > R(\phi',\theta_1)$. This contradicts the fact that $\phi$ solved (\ref{equation:optimization}) subject to  (\ref{equation:sizecontrol}).\\

\item Suppose $1>\alpha=R(\phi,\theta_0) > R(\phi', \theta_0)$ and $R(\phi,\theta_1) \geq R(\phi',\theta_1)$. Consider the test $\phi''$ that rejects for every value of $x$. Such test has Type I and II error given by $(1,0)$. Consider the test
$$\phi'''(x)= \lambda \phi''(x) + (1-\lambda) \phi'(x), \quad \lambda \in [0,1] $$
\noindent This randomized test has Type I error $\lambda + (1-\lambda)R(\phi', \theta_0)$. Since $\alpha \in (R(\phi',\theta_0),1)$, there exists $\lambda(\alpha)>0$ such that:
$$(R(\phi''',\theta_0), R(\phi''',\theta_1))= (\alpha, (1-\lambda(\alpha))R(\phi'(x),\theta_1)).$$
\noindent This contradicts the fact that $\phi$ solved (\ref{equation:optimization}) subject to  (\ref{equation:sizecontrol}).

\item Finally, suppose that $1=R(\phi,\theta_0)>R(\phi',\theta_0)$ and $R(\phi,\theta_1) \geq R(\phi',\theta_1)$. We have already seen that the test
$\phi''$ which rejects for every value of $x$ has $R(\phi'',\theta_0)=1$ and $R(\phi'',\theta_1)=0$. Since $\phi$ solves the minimisation problem for $\alpha=1$, it must be the case that $R(\phi,\theta_1) \leq R(\phi'',\theta_1) = 0$. Then, by assumption $0 \leq R(\phi',\theta_1) \leq R(\phi,\theta_1) \leq 0$. Therefore, $R(\phi',\theta_1) = R(\phi,\theta_1) = 0$. Since $R(\phi',\theta_0)<1$, this implies

\begin{align*}
\int_{\{x \in \textbf{X} | \phi'(x)=0 \}} f(x,\theta_0) dx > 0
\end{align*}
which implies

\begin{align*}
\int_{\{x \in \textbf{X} | \phi'(x)=0 \}} f(x,\theta_1) dx > 0
\end{align*}
which implies $R(\phi',\theta_1)>0$. Contradiction.

\end{enumerate}

\end{proof}

The implication of this proposition is important: if you want to use an admissible test for the simple null against simple alternative then you have no choice but to maximize power subject to a size control constraint. The following proposition tells us how to solve this optimization problem: reject the null hypothesis if the likelihood of the data under the alternative is sufficiently higher than the likelihood under the null.

\begin{proposition} (The Neyman-Pearson Lemma) Consider the  hypothesis testing problem of a simple null $\theta_0$ against a simple alternative $\theta_1$. Define the \emph{likelihood ratio statistic} as
\begin{equation*}
L(x) \equiv \frac{f(x, \theta_1)}{f(x, \theta_0)}
\end{equation*}
and suppose that for each $\alpha \in (0,1)$ there exists $c_{LR}(\alpha)$ such that
\begin{equation*}
P_{\theta_0} (L(X)>c_{LR}(\alpha))=\alpha
\end{equation*}
Then the test:
\begin{equation} \label{equation:NPtest}
\phi(x)= 1_{L(x)>c_{LR}(\alpha)}
\end{equation}
solves:

\begin{equation*}
\phi \in \arg \max_{\phi} (1-R(\phi, \theta_1) )
\end{equation*}
\noindent subject to
$$R(\phi,\theta_0) \leq \alpha $$

\end{proposition}

\begin{proof}
Let $\phi(x)$ be defined as in (\ref{equation:NPtest}). We would like to show the following: if $t(x)$ is another test of level $\alpha$, then $P_{\theta_1} (\phi(x)=1) \geq P_{\theta_1} (t(x)=1) \geq 0$. The prove proceed goes follows. By definition of $\phi$
\begin{equation*}
\phi(x)=1 \quad\quad \text{if}  \quad f(x, \theta_1) > c_{LR}(\alpha) f(x, \theta_0)
 \end{equation*}
Since $t(x) \in [0,1]$ it follows that
\begin{equation*}
\phi(x) - t(x) \geq 0 \quad\quad \text{if}  \quad f(x, \theta_1) > c_{LR}(\alpha) f(x, \theta_0)
 \end{equation*}
 and
 \begin{equation*}
\phi(x) - t(x) \leq 0 \quad\quad \text{if}  \quad f(x, \theta_1) \leq c_{LR}(\alpha) f(x, \theta_0)
 \end{equation*}
 Therefore, the function:
 \begin{equation*}
\left[ \phi(x)-t(x) \right] \left[ f(x, \theta_1) - c_{LR}(\alpha) f(x, \theta_0) \right] \geq 0
 \end{equation*}
 Note then that, for all $x$
 \begin{equation*}
 \left[ \phi(x)-t(x) \right] f(x, \theta_1) \quad \geq  \quad \left[ \phi(x)-t(x) \right] c_{LR}(\alpha) f(x, \theta_0)
 \end{equation*}
 Therefore, integrating with respect to x:
  \begin{equation*}
\int_{\mathbf{X}} \left[ \phi(x)-t(x) \right] f(x, \theta_1)dx \quad \geq  \quad \int_{\mathbf{X}}\left[ \phi(x)-t(x) \right] \lambda f(x, \theta_0) dx
 \end{equation*}
 Note, that the right hand side of the previous equation is exactly the same as the difference in levels (scaled by $\lambda$) of the tests $\phi$ and $t$. Since the Neyman-Pearson test has level $\alpha$ and any other competing test has level at most $\alpha$, then the right-hand side is larger than or equal to zero.  Re-arranging the expression we get:
 \begin{equation*}
 \int_{\mathbf{X}} \phi(x) f(x, \theta_1) \geq  \int_{\mathbf{X}} t(x) f(x, \theta_1)
 \end{equation*}
 And this completes the proof (make sure you understand why these integrals equal the power of the tests)
\end{proof}

\subsection{Testing hypotheses in Parametric Models}

\subsubsection{Generalized Likelihood Ratio Test}

We have just shown that the test that rejects the null hypothesis whenever the likelihood under the alternative is (sufficiently) larger than the likelihood under the null cannot be dominated. To establish this result we assumed that both the null and the alternative were ``simple’’. Unfortunately, most problems we will encounter in econometric practice involve ``composite’’ hypotheses.  \\

\noindent {\scshape Example:} Consider the linear regression model with non-stochastic regressors:
\begin{equation}
Y \sim \mathcal{N}_n(X \beta, \sigma^2 \mathbb{I}_n ).
\end{equation}
Assume (as we did in the last problem set) that both $\beta \in \mathbb{R}^{k}$ and $\sigma^2 \in \mathbb{R}_{+}$ are unknown. Suppose that we are interested in the problem

\begin{equation}\label{equation:Hypothesis}
\textbf{H}_0: \beta=\beta_0 \textrm{ vs. }  \textbf{H}_1: \beta \neq \beta_0.
\end{equation}

\noindent Both the null and the alternative  are composite. To see this, write
\[ \Theta_0 \equiv \{ (\beta, \sigma^2) \: | \: \beta = \beta_0 \}, \quad \Theta_1 \equiv \{ (\beta, \sigma^2) \: | \: \beta \neq \beta_0 \}.\]

\noindent The value of $\sigma^2$ is not specified under the null hypothesis, hence any tuple $(\beta_0, \sigma^2)$ is plausible under the null. Parameters that are not specified under the null hypothesis are usually called \emph{nuisance} parameters.

The \emph{generalized likelihood ratio} statistic provides a general approach for testing hypothesis in parametric models with composite null and alternative hypotheses. The test rejects $\Theta_0$ in favor of the alternative whenever the \emph{likelihood ratio statistic}

\begin{equation}
2\Big [  \max_{\theta \in \Theta_1} \ln f(x | \theta) - \max_{\theta \in \Theta_0} \ln f(x | \theta) \Big ]
\end{equation}
is large enough. The first term denotes the largest value that the ``log-likelihood’’ can achieve under the alternative, and the second term denotes the largest value that the log-likelihood can achieve under the null. Let $\widehat{\theta}_1$ and $\widehat{\theta}_0$ denote the values of the parameter that maximize the likelihood under the alternative and the null, respectively. The generalized likelihood ratio test thus rejects whenever
\[ 2 \ln \left(  f(x | \widehat{\theta}_1) / f(x, \widehat{\theta_0}) \right) \]
is large enough. \\

\noindent {\scshape Likelihood Ratio Test for the Linear Regression Model:} We now compute the generalized likelihood ratio statistic for the testing problem (\ref{equation:Hypothesis}) in the linear regression model. First, we maximize the likelihood under the alternative hypothesis $\beta \neq \beta_0$. This is the same as just maximizing the likelihood over the whole parameter space

\[  \max_{\beta, \sigma^2} f(Y | \beta , \sigma^2). \]

\noindent We solved in the previous problem set and we showed that:

\[ \widehat{\beta}_{\textrm{ML}} = (X’X)^{-1}X’Y, \quad \widehat{\sigma}^2_{\textrm{ML}} = (Y-X\widehat{\beta}_{\textrm{ML}})’(Y-X\widehat{\beta}_{\textrm{ML}})/n. \]

\noindent Thus, we can verify that the largest value of the log-likelihood under the alternative is
\[ \log f(Y | \widehat{\beta}_{\textrm{ML}}, \widehat{\sigma}^2_{\textrm{ML}}) = - \frac{n}{2} \log(2 \pi ) - \frac{n}{2} \log( \widehat{\sigma}^2_{\textrm{ML}}) - \frac{n}{2}. \]

\noindent We now turn to the problem of maximizing the log-likelihood under the null hypothesis:
\[ \max_{\sigma^2} f(Y | \beta_0 , \sigma^2). \]
Algebra shows that the maximizer is

\[\widehat{\sigma}^2_{0} = (Y-X\beta_0)’(Y-X \beta_0)/n,\]

\noindent and the maximized value of the log-likelihood under the alternative corresponds to

\[ \log f(Y | \widehat{\beta}_{0}, \widehat{\sigma}^2_{0}) = - \frac{n}{2} \log(2 \pi ) - \frac{n}{2} \log( \widehat{\sigma}^2_{0}) - \frac{n}{2}. \]
\noindent This means that the likelihood ratio test statistic equals

\begin{equation} \label{equation:GLR}
n  \left( \ln ( \widehat{\sigma}^2_{0} ) -\ln( \widehat{\sigma}^2_{\textrm{ML}} ) \right).
\end{equation}

The \emph{critical value}---the threshold against which the likelihood ratio statistic is compared against---is usually taken as the $1-\alpha$ quantile of $\chi^2$ distribution with degrees of freedom given by the dimension of $\beta_0$. We now explain this approximation for the linear regression model and then present an heuristic derivation of the more general approximation result.

Algebra shows that we can write the likelihood ratio test statistic above as

\[n  \left( \ln \left(  (\widehat{\sigma}^2_{0} - \widehat{\sigma}^2_{\textrm{ML}})/\widehat{\sigma}^2_{\textrm{ML}} +1  \right) \right). \]

\noindent Under the null hypothesis we expect $\widehat{\sigma}^2_0$ and $\widehat{\sigma}^2_{\textrm{ML}}$ not to be very different. Since a standard first-order Taylor appproximation suggests that $\ln(1+x) \approx x$ when $x$ is small, the likelihood ratio test statistic is expected to be approximately equal to

\[ n(\widehat{\sigma}^2_{0} - \widehat{\sigma}^2_{\textrm{ML}})/\widehat{\sigma}^2_{\textrm{ML}}.\]

\noindent Interestingly, in the linear regression model

\[n(\widehat{\sigma}^2_{0} - \widehat{\sigma}^2_{\textrm{ML}}) = (\widehat{\beta}_{\textrm{ML}} - \beta_0 ) (X’X)  (\widehat{\beta}_{\textrm{ML}} - \beta_0 )  = \sigma^2 \frac{1}{\sigma^2} (\widehat{\beta}_{\textrm{ML}} - \beta_0 ) (X’X)  (\widehat{\beta}_{\textrm{ML}} - \beta_0 ).  \]

\noindent Moreover, if the null hypothesis is true

\[(\widehat{\beta}_{\textrm{ML}} - \beta_0 ) = (X’X)^{-1} X’(Y - X \beta_0) \sim \mathcal{N}_k(0, \sigma^2 (X’X)^{-1}),\]

\noindent implying

\[\frac{1}{\sigma^2} (\widehat{\beta}_{\textrm{ML}} - \beta_0 )’ (X’X)  (\widehat{\beta}_{\textrm{ML}} - \beta_0 ) \]

\noindent has the distribution of a $\chi^2$ random variable with $k$ degrees of freedom. If we further assume that when the sample size is large $\sigma^2 / \widehat{\sigma}^2_{\textrm{ML}}=1$, with probability close to one then:

\[ n(\widehat{\sigma}^2_{0} - \widehat{\sigma}^2_{\textrm{ML}})/\widehat{\sigma}^2_{\textrm{ML}} \approx \chi^2_{k}.\]

\subsubsection{The Wald Test}

The approximation result above is more general. Consider the problem
\[\textbf{H}_0: \theta=\theta_0 \textrm{ vs. }  \textbf{H}_1: \theta \neq \theta_0. \]

\noindent The maximized log-likelihood ratio under the alternative is

\[ \ln f( x | \widehat{\theta}_{\textrm{ML}}).\]

\noindent Under the null, $\widehat{\theta}_{\textrm{ML}}$ should be close to $\theta_0$. Thus, an heuristic first-order Taylor approximation of $\ln f(x|\theta_0)$ around $\widehat{\theta}_{\textrm{ML}}$ suggests that

\begin{eqnarray*}
 \ln f(x| \theta_0) &\approx& \ln f( x | \widehat{\theta}_{\textrm{ML}}) + \left( \frac{\partial}{\partial \theta} \ln f( x | \widehat{\theta}_{\textrm{ML}})\right )’ (\theta_0-\widehat{\theta}_{\textrm{ML}}) \\
&+& \frac{1}{2} (\widehat{\theta}_{\textrm{ML}}-\theta_0)’  \left( \frac{\partial^2}{\partial \theta \partial \theta} \ln f( x | \widehat{\theta}_{\textrm{ML}})\right ) (\widehat{\theta}_{\textrm{ML}}-\theta_0).
\end{eqnarray*}

\noindent The term

\[ \left( \frac{\partial}{\partial \theta} \ln f( x | \widehat{\theta}_{\textrm{ML}})\right ) \]

\noindent is the score (the derivative of the log-likelihood) evaluated at $\widehat{\theta}_{\textrm{ML}}$. The F.O.C. defining $\widehat{\theta}_{\textrm{ML}}$ imply this term equals zero. Therefore, the likelihood ratio statistic is approximately equal to
\[2(\ln f( x | \widehat{\theta}_{\textrm{ML}}) - \ln f(x| \theta_0)) \approx (\widehat{\theta}_{\textrm{ML}}-\theta_0)’  \underbrace{\left( -\frac{\partial^2}{\partial \theta \partial \theta} \ln f( x | \widehat{\theta}_{\textrm{ML}})\right )}_{\widehat{\mathcal{I}}: \textrm{ Observed Information Matrix}}  (\widehat{\theta}_{\textrm{ML}}-\theta_0).    \]

\noindent The last term in the equation above is called the Wald test statistic for the null hypothesis $\theta = \theta_0$. Asymptotic theory (which you will cover in the next block of the course) will show that under very general conditions

\[ \widehat{\mathcal{I}}^{1/2}(\widehat{\theta}_{\textrm{ML}}-\theta_0) \sim \mathcal{N}_{\textrm{dim}(\theta)} (0, \mathbb{I}_{\textrm{dim}(\theta)}).\]

\noindent Therefore, the Wald test statistic has approximately the same distribution as a $\chi^2$ random variable with $\textrm{dim}(\theta)$ degrees of freedom. This means that if we let $\chi^2_{\textrm{dim}(\theta),1-\alpha}$ denote the $1-\alpha$ quantile of a $\chi^2_{\textrm{dim}(\theta)}$ random variable then the test that rejects whenever

\[  (\widehat{\theta}_{\textrm{ML}}-\theta_0)’  \widehat{\mathcal{I}} (\widehat{\theta}_{\textrm{ML}}-\theta_0) > \chi^2_{\textrm{dim}(\theta),1-\alpha} \]
will have size of approximately $1-\alpha$. Moreover, the outcome of this Wald test will be approximately the same as that of the Likelihood Ratio Test. \\

\noindent {\scshape Wald Test for the Linear Regression Model:} We derive the Wald test for the problem
\[ \textbf{H}_0: \beta=\beta_0 \textrm{ vs. }  \textbf{H}_1: \beta \neq \beta_0,\]
using the Linear Regression Model with unknown variance. Note that $\sigma^2$ is a nuisance parameter, as it is not specified under the null hypothesis. To construct the Wald test statistic we will need to replace $\sigma^2$ by its ML estimator.

We showed in the previous problem set that the score is given by
\[\begin{pmatrix}
\frac{1}{\sigma^2}X’(Y-X\beta) \\
-\frac{n}{2} \frac{1}{\sigma^2} + \frac{1}{\sigma^4} (Y-X\beta)’(Y-X\beta)
 \end{pmatrix}.\]
 The upper $k \times k$ block of the sample information matrix is then given by the matrix

 \[  \frac{1}{\sigma^2} X’X, \]

\noindent which does not depend on $\beta$, but depends $\sigma^2$. The Wald statistic is thus
\[  \frac{1}{\widehat{\sigma}^2_{\textrm{ML}}} (\widehat{\beta} - \beta_0)’(X’X)(\widehat{\beta} - \beta_0).   \]

\subsubsection{The Score Test}
Finally, we introduce the score test. Once again, consider the problem
\[\textbf{H}_0: \theta=\theta_0 \textrm{ vs. }  \textbf{H}_1: \theta \neq \theta_0. \]
In any parametric model, the ML estimator satisfies the first order condition
\[ \frac{\partial}{\partial \theta} \ln f(x | \widehat{\theta}_{\textrm{ML}})  = 0. \]
An heuristic first-order Taylor approximation thus suggests
\[  \underbrace{\frac{\partial}{\partial \theta} \ln f(x | \theta_0)}_{S(x; \theta_0): \textrm{ Score at $\theta_0$} } \approx \underbrace{\frac{\partial}{\partial \theta} \ln f(x | \widehat{\theta}_{\textrm{ML}})}_{S(x; \theta_0): \textrm{ F.O.C.} } \quad + \underbrace{-\frac{\partial^2}{\partial \theta \partial \theta} \ln f(x | \theta_0)}_{\widehat{I}: \textrm{ observed information} } \left( \widehat{\theta}_{\textrm{ML}} - \theta_0 \right), \]
implying

\[ \left( -\frac{\partial^2}{\partial \theta \partial \theta} \ln f(x | \theta_0) \right)^{1/2} \left( \widehat{\theta}_{\textrm{ML}} - \theta_0 \right)  \approx  \left( -\frac{\partial^2}{\partial \theta \partial \theta} \ln f(x | \theta_0) \right)^{-1/2} S(x; \theta_0).  \]

\noindent If it is true that $\widehat{\mathcal{I}}^{1/2}(\widehat{\theta}_{\textrm{ML}}-\theta_0) \sim \mathcal{N}_{\textrm{dim}(\theta)} (0, \mathbb{I}_{\textrm{dim}(\theta)})$ (an approximation we have used to motivate the Wald Test and to derive the critical value for the Likelihood Ratio Test) then we must have

\[  S(x; \theta_0)’ \left( -\frac{\partial^2}{\partial \theta \partial \theta} \ln f(x | \theta_0) \right)^{-1} S(x; \theta_0) \approx \chi^2_{\textrm{dim}(\theta)} \]

\noindent The score test (with nominal size $\alpha$) rejects if the \emph{score test statistic above} is larger than the $1-\alpha$ quantile of a $\chi^2_{\textrm{dim}(\theta)}$. Note that the score does not require computing the ML estimator, only the F.O.C.

\newpage

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\end{document}


\subsection{Cost model: compression estimators}
\label{sub:estimators}

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The compression learning process is an optimization problem (\ref{subsec:learningprocess:optimizationproblem}) for finding the best compression tree in terms of disk size of the resulting physical columns. Solving the optimization problem requires a way to compare its solutions, i.e. estimating the final size of the compressed data. For this purpose we created a cost model which relies on leaf compression schemes size estimators (DICT, RLE, FOR, no compression) to predict the size of a column if compressed with these methods.

\textbf{Methodology}. The score of a solution (compression tree) is computed through the following methodology: 1) represent the sample data according to the compression nodes; 2) for each column of the new representation, estimate its size if it were compressed with: DICT, RLE, FOR or not compressed at all; 3) choose the smallest size among those. The result of this process is the smallest size of the new data representation if it were compressed with leaf compression schemes.

\textbf{System assumptions}. The size on disk of a compressed column depends on: 1) implementation of the compression method; 2) characteristics of the underlying database system. The former is described in the next sections as part of the estimator implementations. For the latter we defined some assumptions of the underlying system as follows:\\
a) \textit{data types}: strings are stored with null terminator, therefore their size is given by the number of characters + 1. For all other data types we consider the sizes used by Ingres Vectorwise \cite{zukowski2012vectorwise,actianingres}. \\
b) \textit{null handling}: we consider the same approach  used by Vectorwise \cite{zukowski2012vectorwise}: do not store null values, instead, keep track of their positions using a bitmap. This results in 1 additional bit for every attribute (for nullable columns).\\
c) \textit{exception handling}: we consider a whitebox approach: for each logical column store exceptions on a separate physical nullable column. The exception column has the same datatype as the logical column.

An additional assumption that we make about the underlying system is that it supports block-level compression, i.e. every block of data is compressed independently. This allows different compression schemes to be used on the same column, enabling the possibility to exploit local data characteristics.

\subsubsection{Generic compression estimator}
\label{subsub:estimator:generic}
A compression estimator takes as input a column and two samples of data (\textit{train} and \textit{test} sample) and outputs the estimated size of the compressed column. The result can be either the size of the compressed sample or the size extrapolated to the total size of the block or column. The latter requires an additional parameter specifying the total number of rows in the full data.

The estimated size has 4 components (exemplified for Dictionary encoding):\\
1) \(size_{metadata}\): size of the metadata (the dictionary itself)\\
2) \(size_{data}\): size of the compressed data (the dictionary ids)\\
3) \(size_{ex}\): size of the exceptions  (the values that are not in the dictionary)\\
4) \(size_{null}\): size required to keep track of the null values. Since exceptions are stored on a separate nullable column, the \textit{nulls} size is implicitly increased.

The final estimated size of the test sample is the sum of the 4 components:
\begin{equation}
\label{eq:estimators:sizesample}
size_{sample} = size_{metadata} + size_{data} + size_{ex} + size_{null}
\end{equation}

This result gives the size of the \textit{test} sample only. It can be extrapolated to the full size of the block or entire column as follows:
\begin{equation}
\label{eq:estimators:sizefinal}
size_{full} = size_{sample} \times \frac{count_{full}}{count_{sample}}
\end{equation}
where:
\begin{itemize}
    \item[] \(count_{sample}\) = total number of values in the test sample
    \item[] \(count_{full}\) = total number of values in the full block or column
\end{itemize}

The size estimation process works in two phases:\\
1) \textit{training}: the estimator analyzes the \textit{train} sample and generates the metadata needed for compression (e.g. \nameref{subsub:estimator:dict} generates the dictionary, \nameref{subsub:estimator:for} determines the reference value and the number of bits needed to store the differences).\\
2) \textit{testing}: the estimator simulates the compression of the \textit{test} sample using the metadata resulted from the \emph{training} phase and outputs an estimated size.

The two-phase estimation process is used to avoid overly-optimistic results: metadata generated based on the \emph{train} sample is perfectly optimized for that sample (e.g. in FOR all differences will fit in the number of bits chosen to represent them). Depending on the implementation of each compression estimator, this would lead to a reduced number of exceptions or even no exceptions at all. Therefore, the \textit{test} sample is used to provide new data for size estimation. It produces exceptions and more realistic results. This approach simulates the compression process of real database systems, where the compression metadata is created based on a sample and then applied on a full block of data or even on the entire column.

The next sections describes 4 compression estimators used in the learning process. All computed sizes will be in bytes.

% ------- no compression ------- %

\subsubsection{No compression estimator}
\label{subsub:estimator:nocompression}

The \nameref{subsub:estimator:nocompression} predicts the size of the input column stored without compression. The \textit{training} phase is not relevant since it does not generate any compression metadata. The estimation is performed in the \textit{testing} phase, based on the size on disk of the column data type. The size components are computed as follows:

\(size_{metadata}\) is 0, since there is no compression metadata

\(size_{ex}\) is 0, since there are no exceptions

\(size_{null}\) is 1 bit for every value in the sample:
\begin{equation}
\label{eq:estimators:nocompression:sizenull}
size_{null} = \frac{count_{sample}}{8}
\end{equation}
where:
\begin{itemize}
    \item[] \(count_{sample}\) = total number of values in the test sample
\end{itemize}

\(size_{data}\) is given by the total size of the non-null values in the sample. It depends on the data type of the column as follows:
\begin{equation}
\label{eq:estimators:nocompression:sizedata}
size_{data} =
\left\{
\begin{array}{ll}
    \sum_{v \neq \mathit{null}} \mathit{len}(v) + 1 & \mbox{if } \mathit{datatype} = \verb|VARCHAR|\\
    count_{notnull} \times size_{datatype} & \mbox{else}
\end{array}
\right.
% \frac{count_{sample}}{8}
\end{equation}
where:
\begin{itemize}
    \item[] \(count_{notnull}\) = total number of non-null values in the test sample
    \item[] \(size_{datatype}\) = size on disk of the column data type
\end{itemize}

% ------- dictionary ------- %

\subsubsection{Dictionary estimator}
\label{subsub:estimator:dict}

The \nameref{subsub:estimator:dict} predicts the size of the input column as compressed with Dictionary encoding. Besides the two samples, it receives an additional parameter: \(size_{max}\)---maximum size of the dictionary (in bytes). It only applies to \verb|VARCHAR| columns and therefore the exception column will also be \verb|VARCHAR|.

The \nameref{subsub:estimator:dict} is similar to the \nameref{subsec:pd:dict} pattern detector defined in \ref{subsec:pd:dict}. It builds the dictionary and handles exceptions in the same way. Optimizing the dictionary based on a maximum size (in bytes) also works for the estimator, since we use it to compare different ways of compressing a column and the dominant factor here is the nature of the data, not the optimization of the compression scheme.

% Dictionary encoding only produces good results on columns that have a small number of unique values. However, it is hard to reliably quantify this property when analyzing only a sample of the data. Moreover, the distribution of unique values may be skewed, with only a few values with high frequency and a long tail of low frequency values. For the purpose of our estimator, we addressed this issue by enforcing a maximum dictionary size and only keeping the most common values in the dictionary. This approach is also suitable if blocks of data are compressed independently: dictionaries need to be small as they are assigned per block. There are other (possibly better) ways of optimizing the dictionary values and size. However, they are out of the scope of our estimator, since we use it to compare different ways of compressing a column and the dominant factor here is the nature of the data, not the optimization of the compression scheme.

\textbf{Training phase.} The dictionary (metadata) is built during the \textit{training} in same way it is done for the \nameref{subsec:pd:dict} pattern detector (\ref{subsec:pd:dict}): Step-1: create the histogram of all the values in the \textit{train} sample. Step-2: select as many values from the histogram in decreasing order of their number of occurrences such that their total size is lower or equal to the maximum size of the dictionary (\(size_{max}\)). The dictionary is stored as an array containing the selected values. The indices in the array represent the dictionary ids used to encode the values.

\(size_{metadata}\) is given by the total size of the values in the dictionary. Additionally, the number of bits required to store a dictionary id is computed as follows:
\begin{equation}
\label{eq:estimators:dict:bitsid}
bits_{id} = \lceil \log_2 (count_{entries}) \rceil
\end{equation}
where:
\begin{itemize}
    \item[] \(count_{entries}\) = number of values in the dictionary
\end{itemize}

\textbf{Testing phase}. The \textit{testing} phase estimates the size of the compressed column by going through each value in the \textit{test} sample and checking if it is present in the dictionary. The following variables are updated in this process: 1) \(count_{valid}\): number of values that are found in the dictionary; 2) \(size_{ex}\): size of the exceptions (values that are not found in the dictionary).

\(size_{data}\) is computed as follows:
\begin{equation}
\label{eq:estimators:dict:sizedata}
size_{data} = \frac{count_{valid} \times bits_{id}}{8}
\end{equation}

\(size_{ex}\) is computed by summing the size of all exceptions.

\(size_{null}\) is determined by the number of resulting physical columns: one for compressed data (dictionary ids) and one for exceptions:
\begin{equation}
\label{eq:estimators:dict:sizenull}
size_{null} = \frac{2 \times count_{sample}}{8}
\end{equation}
where:
\begin{itemize}
    \item[] \(count_{sample}\) = total number of values in the test sample
\end{itemize}

% ------- run length encoding ------- %

\subsubsection{Run Length Encoding estimator}
\label{subsub:estimator:rle}

The \nameref{subsub:estimator:rle} predicts the size of the input column as compressed with RLE. The samples are constructed with consecutive ranges of values such that RLE is triggered (see \ref{subsec:eval:methodology:sampling}~\nameref{subsec:eval:methodology:sampling} for more details). Even though RLE can be applied to any data type, we limited the scope of our estimator to numeric columns. The other data types are either compressed with Dictionary encoding (\verb|VARCHAR|) or are very rare in the Public BI benchmark and do not present compression opportunities. We use the following terminology:\\
1) \textit{run} value = a data value that is repeated on consecutive rows.\\
2) \textit{length} value = the number of consecutive occurrences of a \textit{run} value

\textbf{Training phase.} RLE metadata is composed of: 1) the number of bits needed to represent the \textit{run} values (\(bits_{run}\)) and 2) the number of bits needed to represent the \textit{length} values (\(bits_{length}\)). These values are determined by scanning the \textit{train} sample and computing all the \textit{runs} and \textit{lengths}. \(bits_{run}\) is given by the \textit{run} value of maximum size and \(bits_{length}\) is given by the maximum \textit{length}. \(bits_{run}\) also depends on the column data type representation.

\(size_{metadata}\) is between 8 and 24 bytes---the size of 2 numbers: \(bits_{run}\) and \(bits_{length}\)---depending on the data types used to store them.

\textbf{Testing phase.} The \textit{testing} phase scans all the values in the \textit{test} sample and creates (\textit{run}, \textit{length}) pairs as follows: 1) if a \textit{run} value cannot be represented on \(bits_{run}\) bits: mark it as exception and skip it; 2) if a \textit{length} exceeds the maximum value that can be represented on \(bits_{length}\) bits: end the current run at this length and start a new run. The following variables are updated in this process: 1) \(count_{valid}\): the number of (\textit{run}, \textit{length}) pairs resulted from the scanning process; 2) \(count_{ex}\): the number of exceptions as defined above.

\(size_{data}\) and \(size_{ex}\) are computed as follows:
\begin{equation}
\label{eq:estimators:rle:sizedata}
size_{data} = \frac{count_{valid} \times (bits_{run} + bits_{length})}{8}
\end{equation}
\begin{equation}
\label{eq:estimators:rle:sizeex}
size_{ex} = count_{ex} \times size_{datatype}
\end{equation}
where:
\begin{itemize}
    \item[] \(size_{datatype}\) = size on disk of the column data type
\end{itemize}

\(size_{null}\) depends on the number of physical columns---one compressed data column (\textit{run} and \textit{length} are stored together) and one exception column---the same as in the case of \nameref{subsub:estimator:dict} (Equation \ref{eq:estimators:dict:sizenull}).

% ------- frame of reference ------- %

\subsubsection{Frame of Reference estimator}
\label{subsub:estimator:for}

The \nameref{subsub:estimator:for} predicts the size of the input column as compressed with FOR. It only applies to numeric columns.

\textbf{Training phase.} FOR metadata is composed of: 1) the \textit{reference} value and 2) the number of bits needed to store the differences (\(bits_{\mathit{diff}}\)). In our implementation we chose the \textit{reference} to be the smallest value in the \textit{train} sample. \(bits_{\mathit{diff}}\) is given by the maximum difference size, which depends on the data type representation.

\(size_{metadata}\) is between 8 and 24 bytes---the size of the reference and the size of \(bits_{\mathit{diff}}\)---depending on the data types used to store them.

\textbf{Testing phase.} The testing phase computes all the differences between the values in the \textit{test} sample and the \textit{reference} and filters the ones that can be represented on \(bits_{\mathit{diff}}\) bits. The following variables are updated in this process: 1) \(count_{valid}\): the number of differences that fit in \(bits_{\mathit{diff}}\); 2) \(count_{ex}\): the number of exceptions (values that give differences larger than \(bits_{\mathit{diff}}\)).

\(size_{data}\) is computed as follows:
\begin{equation}
\label{eq:estimators:for:sizedata}
size_{data} = \frac{count_{valid} \times bits_{\mathit{diff}}}{8}
\end{equation}

\(size_{ex}\) is the same as in the case of \nameref{subsub:estimator:rle} (Equation \ref{eq:estimators:rle:sizeex}).

\(size_{null}\) depends on the number of physical columns---one compressed data column (differences) and one exception column---the same as in the case of \nameref{subsub:estimator:rle} and \nameref{subsub:estimator:dict} (Equation \ref{eq:estimators:dict:sizenull}).

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