\chapter{Coprime action and $p$-constraint}
\section {Basic Results}
{\bf Definition 1:} A group of automorphisms $A$ of a group $P$ stabilizes a chain
$1=P_n \subseteq P_{n-1} \subseteq \ldots \subseteq P_0 = P$ if
$[A,P_i] \subseteq P_{i+1}$.
\\
\\
{\bf Theorem 1:} If $P$ is a $\pi$ group with chain stabilized by
$A$ then $A$ is a $\pi$ group.
\begin{quote}
\emph{Proof:} Suppose $a \in A$ is a $\pi'$ automorphism.  Proof is by induction on the
length of the chain and so we can assume $[a,P_1]=1$.
$x^a =xy, y \in P_1$.  So, $x^{a^{|a|}}= x y^{|a|}=x$, so $y=1$ and
$[a,P]=1$.
\end{quote}
{\bf Theorem 2:} If $A$ is a $\pi'$ group of automorphisms on a $\pi$ group $P$ with
$[P,A,A]=1$ then $[P,A]=1$.
\begin{quote}
\emph{Proof:}  $A$ stabilizes
$[P,A,A] \subseteq [P,A] \subseteq P$.
\end{quote}
{\bf Theorem 3:}
Let $A$ be a $\pi'$ group of automorphisms of a $\pi$ group $P$.  Let $Q$ be
an $A-$invariant normal subgroup of $P$.  Then $C_{P/Q}(A)= (C_P(A) Q)/Q$.
\begin{quote}
\emph{Proof:}   $C_P(A)Q/Q \subseteq C_{P/Q}(A)$.  Suppose $xQ$ is a subset in $P$ fixed by
$A$.  $QA$ acts transitively on the set $xQ$.  Let $A_1$ be the point stabilizer.
$|A_1|=|QA|/|xQ|=|A|$.  By Schur-Zassenhaus, $A$ and $A_1$ are conjugate, so $\exists y \in xQ:
y^A=y$.
\end{quote}
{\bf Theorem 4:} If $P$ is a $\pi$ group, $A$ is a $\pi'$ group, $P= [P,A] C_P(A)$.
\begin{quote}
\emph{Proof:}
$[P,A] \subseteq P$ and $A$ centralizes $P/[P,A]$.
\end{quote}
{\bf Lemma A:} If the action of $A$ on $G$ is co-prime and $U$ is an $A$-invariant subgroup of $G$ with
$(Ug)^A=Ug$, then $\exists c \in C_G(A)$: $Ug = Uc$.
\begin{quote}
\emph{Proof:}  $[a, g^{-1}] \in U$ so $A^{g^{-1}} \leq AU$.  Both $A$ and $A^{g^{-1}}$ are $U$ complements in
$AU$ so by S-Z, $\exists u \in U$: $A^{g^{-1}} = A^u$.  Put $c= ug$.  $c \in N_{AG}(A) \cap Ug$, so
$[A,c] \subseteq A \cap G = 1$.
\end{quote}
{\bf  Theorem:} Let $N$ be an $A$-invariant normal subgroup of $G$ and
let $A$ act co-primely on $G$ then (i)
$C_{G/N}(A)= C_G(A)N/N$ and
(ii) if $A$ acts trivially on $N$ and $G/N$ then $A$ acts trivially on $G$.
\begin{quote}
\emph{Proof:}  $C_G(A)N/N \subseteq C_{G/N}(A) $ is obvious.
If $Ng$ is in $C_G(A)N/N$, $(Ng)^A=Ng$ and by
the lemma, $\exists c \in C_G(A): Ng=Nc$, proving (i).  ii follows directly from the Lemma.
\end{quote}
{\bf Alternate proof of 2, 4:}
\begin{quote}
\emph{Proof:}
$G/[G,A] = C_{G/[G,A]}(A) = C_G(A)[G,A]/[G,A]$, proving (4).
Let $g \in G$ then $g = hk, h \in [G,A], k \in C_G(A)$ by the above.
Since $[xy, z]= [x,z]^y [y,z]$, for $a \in A$,
$[g,a]= [hk,a] = [h,a]^k [k,a]$.  $[k,a]=1$ since $k \in C_G(A)$.  Thus
$[g,a] = [h,a]^k \in [G,A,A]$.
\end{quote}
{\bf Alternate proof of Thompson:}  Let $A= P \times Q$ act on a $p$-group $G$ and suppose $C_G(P) \leq C_G(Q)$.  Then
$Q$ acts trivially on $G$.
\begin{quote}
\emph{Proof:}  $C_U(P) \leq C_U(Q)$ for all $A$-invariant subgroups, $U$.  By induction, we can assume
$[U,Q]=1$ for all proper subgroups.  $[G,P,Q]=1$ and $[P, Q, G] =1$ so $[Q, G, P]=1$.  Thus
$[Q,G] \leq C_G(P) \leq C_G(Q)$ and $[G,Q,Q] = 1$.  By the previous result, $[G,Q]=[G,Q,Q]=1$.
$G/[G,A] = C_{G/[G,A]}(A) = C_G(A)[G,A]/[G,A]$.
\end{quote}
{\bf Theorem 5:} $P$ is an abelian $\pi$ group, $A$ is a $\pi'$ group.  $P= [P,A] \oplus C_P(A)$.
\begin{quote}
\emph{Proof:}
$\theta= {\frac 1 {|A|}} \sum_a a$.  For $a \in A$, $\theta a = a \theta = \theta$ and
$\theta^2= \theta$. $\theta P \oplus ker(\theta)= P$.  $C_P(A) \subseteq \theta P$ and
$\theta([x,a]) =0$, so $[P,A] \subseteq ker(\theta)$.  This shows $C_P(A) \cap [P,A] = \emptyset$
and the result follows from the earliler result that $P= [P,A] C_P(A)$.
\end{quote}
{\bf Theorem 6:}
If $\phi$ is a $p'$-automorphism of a $p$-group, $P$, which induces the identity on
$P/ \Phi(P)$ then $\phi= 1$.
\begin{quote}
\emph{Proof:}
Let $H= \Phi(P), {\overline P}= P/H, |{\overline P}|= p^r, |H|= p^m$.
$|P|= p^{m+r}$ and for any subset $Y= \{ y_i, 1 \le i \le n \} \subseteq P$,
$P= \langle P, Y \rangle$ iff $P= \langle Y \rangle$.  Note that ${\overline P}$ and
hence $P$ cannot be generated by $<r$ elements.  Let $\{ x_i \}$ be a minimal generating
set for $P$ and $x_i'= h_i x_i$ be another such generating set.  Let $\phi$ be as set forth in the
statement of the theorem.  Finally, let ${\cal M}$ be the collection of all such minimal generating
sets.  $\phi$ fixes each coset  $H x_i'= H x_i$ and $\phi(x_i')= h_i x_i$ so $\phi$
indusces a permutation representation on ${\cal M}$.  The cycle length, $s$, of any cycle in this
permutation representation of $\phi$ must divide $t= | \phi |$.  Suppose some such cycle
had length $s<t$.  $\phi_1= \phi^s$ fixes the elements of this cycle and so fixes some
minimal generating set.  But then $\phi_1$ fixes evely element of $P$ contrary
to the fact that $t>s$.  Since ${\cal M}$ decomposes as a product of disjoint cycles,
$t \mid |{\cal M}| = p^{mr}$.  Since $(t,p)= 1$, $t=1$ and $\phi= 1$ on $P$.
\end{quote}
{\bf $P \times Q$ lemma:} Let $A = P \times Q$, $P$ a $p-$group, $Q$ a $p'$-group, act
on a $p$-group $G$.  If $C_G(P) \le C_G(Q)$ then $Q$ acts trivially on $G$.
\begin{quote}
\emph{Proof:}
$C_U(P) \le C_U(Q)$ for all $A$-invariant subgroups $U$.  By induction, $[U,A]=1$ if
$U < G$.  Now $[G,P] < G$ so $[G, P, Q] = 1$ and since $[P, Q]=1$,
$[P, Q, G]=1$ so by the three subgroups lemma $[Q, G, P]=1$ so
$[Q,G] \le C_G(P) \le C_G(Q)$ and $[G, Q, Q]=1$ hence $[G,Q]=1$.
\end{quote}
{\bf Theorem 7:} Suppose $M$ is a $p$-group and $C_M(P) \le C_M(Q)$.
Then $Q$ acts trivially on $M$.
\begin{quote}
\emph{Proof:}  This is just a restatement of the $P \times Q$ Lemma.
\end{quote}
{\bf Application of $P \times Q$:}  Let
$P \in p(X)$, $M=O_p(X)$, $Q= O_{p'}(N_X(P))$.  $[P, Q]=1$ so
$P \times Q$ acts on $M$.
Next, $[C_M(P), Q] = 1$ because $[C_M(P), Q] \subseteq M$ and $[C_M(P), Q] \subseteq Q$, since
$C_M(P)$ normalizes $P$ so $C_M(P)$ normalizes $Q=O_{p'}(N_X(P))$; as a result, we have,
$[C_M(P), Q] = M \cap Q = 1$..
Let $x \in C_M(P)$ then $[x, Q] = 1$ so $x \in C(Q)$ and $x \in M$ so
$x \in C_M(Q)$.  Thus $C_M(P) \subseteq C_M(Q)$ and the $P \times Q$ lemma applies, that is, $[M,Q]=1$.
If $X$ be solvable, we can use this to show $O_{p'}(N_X(P)) \subseteq O_{p'}(X)$, since $C_X(F(X)) \subseteq F(X)$.
\\
\\
{\bf Theorem 8:}
Suppose $X$ acts faithfully on an Abelian group $A$ and $X= \bigcup_{i=0}^t X_i$ with
$X_i \cap X_j =1$ for $i \ne j$ then $C_A(X_i) \ne 1$ for some $i$ or
$a^t=1, \forall a \in A$.
\begin{quote}
\emph{Proof:}
Let $\sigma_i(a)= \prod_{x \in X_i} a^x$ and
$\sigma(a)= \prod_{x \in X} a^x$.
If the first condition doesn't hold for $i$, $(\sigma_i(a))^x = \sigma_i(a)$ for
all $x \in X_i$, so $\sigma_i(a) = 1$.  Similarly, $\sigma(a)$ is fixed
by all elements of $X$ and since $X$ acts faithfully, $\sigma(a) = 1$.
If $\sigma_i(a)=\sigma(a)= 1, \forall a, i$ then
$\prod_{i=0}^t \sigma_i(a)= \sigma(a) a^t$, so $a^t=1$.
\end{quote}
{\bf Application:} $X= {\mathbb Z}_p \times {\mathbb Z}_p$, $|A| \ne 0 \jmod{p}$.
\\
\\
{\bf Theorem 9:}
If $A$ is abelian and $A$ acts on $G$, $(|A|, |G|)=1$ then
$G= \langle C_G(A_0): A/A_0 \textnormal{ is cyclic} \rangle $.
\begin{quote}
\emph{Proof:}
By induction on $|AG|$.  We may assume $A$ acts faithfully otherwise $C_A(G) \ne 1$ and
the result holds.
\\
\\
\emph{Claim:} Under the hypothesis, $\exists P \in S_p(G): P$ is
$A$-invariant.
\\
\emph{Proof of claim:}  By Frattini, if $P \in S_p(G) \rightarrow AG=GN_{AG}(P)$.
$\exists A_1 \subset N_{AG}(P): A_1 \cap G=1$ and $A_1 N_{AG}(P)$ so $\exists g \in N_{AG}(P):
A_1^g=A$ so $A \subseteq N_G(P)$.   This proves the claim.
\\
\\
If $G$ is not a $p$-group, $\exists P \in S_p(G)$ for each $p$ which are $A$-invariant
so $ \langle C_G(A_0) \rangle \supseteq C_P(A_0)=P$ and we are done.
Assume $G$ is a $p$-group.  If $A$ is cyclic, the result holds trivially.  Suppose
$A_0 < A$ and $A_0= {\mathbb Z}_p \times {\mathbb Z}_p$ and ${\overline P}= P/\Phi(P)$ then
$A$ acts on ${\overline P}$.  By the previous result,
$\exists a \in A_0^{\#}: C_{\overline P}(a) \ne 1$.  Put
${\overline P}_0= C_{\overline P}(a)$.
If ${\overline P}_0={\overline P}$, $\forall a: [P, a] \subseteq \Phi (P)$ which contradicts
the faithfulness hypothesis.  So ${\overline P}_0 < {\overline P}$, ${\overline P}_0$ is
$A$-invariant.  Since ${\overline P}$ is elementary abelian, $\exists {\overline P}_1$
also $A$-invariant: ${\overline P}= {\overline P}_0 \times {\overline P}_1$.  Thus
$P= P_0 P_1$, $P_0 \cap P_1 \subseteq \Phi (P)$ and by induction,
$P_i= \langle C_{P_i}(A_0): A/A_0 \textnormal{ is cyclic } \rangle $.
\end{quote}
{\bf Theorem 10:}  Suppose $N \lhd G$, $H < G$, ${\overline G}= G/N$.  If $(|N|, |H|)=1$
and either is solvable then
(a) $N_{{\overline G}}({\overline H}) = {\overline {N_G(H)}}$ and
(b) $C_{{\overline G}}({\overline H}) = {\overline {C_G(H)}}$.
\begin{quote}
\emph{Proof:}
Note that ${\overline {N_G(H)}} \subseteq N_{\overline G}({\overline H})$ which is the
inverse image of $N_G(HN)$.  We want to show $N_G(HN)= N_G(H)N$.
Let $g \in N_G(HN)$.  $H, H^g$ are complements to $N$ in $HN$ so by Schur-Zassenhaus,
$H^{gn}=H, n \in N$ and $gn=m \in N_G(H)$ and thus $g \in N_G(H)N$.  Let $K$ be the
inverse image in $G$ of $C_{\overline G}({\overline H})$ and $K \supseteq N$.
We have $K \subseteq KN \subseteq N_G(H)N$ and so by Dedekind, $K=N N_N(H)$.
Since $[ {\overline K}, {\overline H}]= 1$, $[K, H] \subseteq N$ and
$[N_K(H), H] \subseteq N \cap H = 1$.  So $N_K(H) \subseteq C_G(H)$ and
$K \subseteq C_G(H)N$ and therefore,
$C_{\overline G}({\overline H})= {\overline K} \subseteq {\overline {C_G(H)}}$.  The
opposite inclusion obviously holds.
\end{quote}
\section {$p$-constraint}
{\bf Definition:} $G$ is $p$-constrained if $O_{p'}(G)=1$ and $C_G(O_p(G)) \leq O_p(G)$ (So
$O_p(G)=F(G)$).
\\
\\
{\bf Theorem 11:}
Suppose $G$ is solvable and $O_{p'}(G)=1$ then $C_G(O_p(G))) \subseteq O_p(G)$.
\begin{quote}
\emph{Proof:}
$O_p(G)$ is nilpotent so $O_p(G) \subseteq F(G)$.  If $Q \in S_q(F(G))$ then
$Q \; char \; F(G)$ so $Q \lhd G$.  Thus $Q=1$ for $q \ne p$ by hypothesis and
$F(G)$ is a $p$-group and $F(G)=O_p(G)$.  Now the result follows since $C_G(F(G)) \subseteq F(G)$.
\end{quote}
{\bf Note:} $G$ acts on $F(G)$ by conjugation so $G/{\mathbb Z}(F(G)) \rightarrow Aut(F(G))$.
\\
\\
{\bf Theorem 12:}  If $G$ is $p$-constrained, $p \in p(G)$ and $P \subseteq G$ then
(1) $O_{p'}(N_G(P)) \subseteq O_{p'}(G)$ and
(2) $N_G(P)$ is $p$-constrained.
\begin{quote}
\emph{Proof:}
(1) Let ${\overline G}= G/O_{p'}(G)$, then, as above,
$N_{{\overline G}}({\overline P}) = {\overline {N_G(P)}}$.
${\overline {O_{p'}(N_G(P))}} \subseteq O_{p'}({\overline {N_G(P)}}) =
O_{p'}(N_{\overline G}({\overline P}))$, so to prove (1), we can assume $O_{p'}(G) = 1$.
Let $Q= O_{p'}(N_G(P))$ and $M= O_p(G)$.  $[P, Q] \subseteq P \cap Q = 1$ and so
$PQ= P \times Q$ acts on $M$.  $[C_M(P),Q] \subseteq M \cap [N_G(P), Q] = 1$ so
by the $p \times q$ lemma, $[M, Q] = 1$.   Hence
$Q \subseteq C_G(M) \subseteq M$ (by $p$-constraint) and $Q = 1$.
\\
(2)
$N_{{\overline G}}({\overline P}) = {\overline {N_G(P)}}= N_G(P)/(N_G(P) \cap O_{p'}(G))$.
By (1), $O_{p'}(N_G(P)) = N_G(P) \cap O_{p'}(G)$ and so
$N_{\overline G}({\overline P})= N_G(P)/O_{p'}(N_G(P))$ so we
may assume $O_{p'}(G) = 1$.
Put $M_1 = O_p(N_G(P))$ and $M=O_p(G)$.  It suffices to show that $C_G(M_1 )$ is a $p$-group.
Let $Q \in S_q(C_G(M_1 )), q \ne p$.  Since $P \lhd N_G(P)$ and $P \subseteq M_1$ and
$P \times Q$ acts on M.  $C_M(P)= O_p(G) \cap C_G(P) \subseteq C_G(P)$ and therefore
$C_M(P) \subseteq O_p (N_G(P))$.  $O_p (C_G(P)) \lhd N_G(P)$ so
$O_p(C_G(P)) \subseteq O_p(N_G(P))= M_1$ and $C_M(P) \subseteq M_1$.  Since
$Q \subseteq C_G(M)$, we have $[C_M(P), Q] = 1$, thus $[M, Q]= 1 $, $Q = 1$, $C_G(M)$ is
a $p$-group.
\end{quote}
{\bf Theorem 13:}  Suppose $G$ is $p$-constrained, $P \in S_p(G)$ and $Q$ is a $P$-invariant
$p'$-subgroup of $G$ then $Q \subseteq O_{p'}(G)$.
\begin{quote}
\emph{Proof:}
Reduce to $O_{p'}(G) = 1$ and let $M= O_p(G)$ so $M \subseteq P$.
$[M, Q] \subseteq M \cap [P, Q] \subseteq M \cap Q = 1$ and $Q = 1$.
\end{quote}
{\bf Theorem 14:}
If $P \in p(G)$ with $N_G(P)$ $p-$constrained then $C_G(P)$ is also $p-$constrained.
\begin{quote}
\emph{Proof:}
$O_{p'}(N_G(P)) = O_{p'}(C_G(P))$ so by considering factor groups, we may assume
$O_{p'}(N_G(P))= 1$.  Let $N= N_G(P)$, $C= C_G(P)$, $Q= O_p(C)$, $R= O_p(N)$.
$Q \; char \; C \lhd N$ and $Q \subseteq R \cap C$, so $Q= R \cap C$.
$R \cap C \lhd C$ and $ R \cap C \subseteq Q$ so $Q = R \cap C$.
Let $x \in C_C(Q)$ be a $p'$ element.   $[x, R] \subseteq R \cap C = Q$,
so $X$ stabilizes the chain $R \supseteq Q \supset 1$ and so $x \in C_N(R) \subseteq R$ as
$N$ is $p$-constrained.  Thus $C_C(Q)$ is a $p$-group and normal in $C$ hence $C_C(R) \subset Q$.
\end{quote}
{\bf Example:} $G$ solvable and $P_1, \ldots, P_n$ a Sylow system.  Let $Q_i= O_{p'}(P_1P_2)$
then $Q_3, \ldots, Q_n$ is a Sylow system for $O_{p'}(G)$.
\\
\\
{\bf Theorem 15:}  Suppose $V$ is a non-cyclic abelian $r$-group of operators on a $p$-constrained
$r'$ group $X$.  For $r \ne p$ then $\bigcap_{v \in V^{\#}} O_{p'}(C_X(v)) \subseteq O_{p'}(X)$.
$p'$-subgroup of $G$ then $Q \subseteq O_{p'}(G)$.
\begin{quote}
\emph{Proof:}
Let ${\overline X}= X/O_{p'}(X)$ be $V$-invariant.
$C_{\overline X}({\overline V}) = {\overline {C_{X}( V)}}$.  We may asssume
$O_{p'}(X) = 1$.  Let
$M = \langle C_M (V_0 ): V/V_0 \rangle = \langle C_M(v), v \in V^{\#} \rangle $
where $V/V_0$ is cyclic.
The last equality holds since $V$ is non-cyclic.
$C_M(v) = M \cap C_X(v) \lhd C_X(v)$ so $C_M(v) \subseteq O_p(C_V(X))$.
Now, $Q \subseteq O_{p'}(C_V(X))$ and $[C_M(V), Q] =1, \forall v \in V^{\#}$ therefore
$[M, Q] = 1$ and so $Q = 1$.
\end{quote}
{\bf Theorem 16:}
If $G$ is solvable, $C_G(F(G)) \subseteq F(G)$.
\begin{quote}
\emph{Proof:}
Let $Z = {\mathbb Z}(F(G))$ and
$N/Z$ be a minimal normal subgroup of $C_G(F(G))/Z$ to obtain a contradiction.
\end{quote}
{\bf Theorem 17 (Another version of the $P \times Q$ Lemma):}
If $P$ is a $p$-group and $Q$ is a $p'$ group, and $P \times Q$ acts on a $p$-group,
$M$ then $[M,Q] = 1$.
\begin{quote}
\emph{Proof:}
For the semi-direct product $(P \times Q)M$ then $P_1= PM$ is $Q$-invariant.  Put
$P_0=P C_M(P)$.  $P_0 \lhd \lhd P_1$, $[P_0, Q]= 1$ so $[M, Q] = 1$.
\end{quote}
{\bf Theorem 18:}
Suppose $X$ acts on $A$, $(|X|, |A|)= 1$, $A$ abelian.  Then
(1) If $A_0 \subseteq A$ is an $X$-invariant direct factor of $A$, $\exists A_1 \subseteq A$:
$A_1$ is $X$ invariant and $A= A_0 \times A_1$; and,
(b) $A= C_A(X) \times [A,X]$.
\begin{quote}
\emph{Proof:}
Let $f: A \rightarrow A_0$ be any projection.  $\exists i: |X|i= 1 \jmod{|A|}$.
Put $f_1(a)= \prod_{x \in X} [f(a^x)^{x^{-1}}]^i$.
$f_1$ is an endomorphism, $im(f_1) \subseteq A_0$, if $a_0 \in A_0, f(a_0) =a_0$ and
$f_1^2= f_1$
so $A= A_0 \times ker(f_1)$.
$f_1(a^y)= (f_1(a))^y, \forall a \in A, y \in X$ so $ker(f_1)$ is $A$-invariant which proves (a).
For (b), let $f(a)= (\prod_{x \in X} a^x)^i$ then
$f(a^y)= f(a)^y$ so $im(f) \subseteq C_A(X)$.   If $a \in C_A(X)$ then
$f(a)=a$ hence $C_A(X)$ is a direct factor of $A$.
$A= C_A(X) \times A_1$, $A_1$, $X$-invariant.  $[A, X] \subseteq A_1$ and
$A_1= [A, X] C_{A_1}(X)$  so $[A, X] \subseteq A_1 \subseteq [A_1 , X] \subseteq [A, X]$.
\end{quote}
{\bf Theorem 19:}   Suppose $G$ is an $X$-group and $(|[G, X]|, |X|) = 1$.
(1) $[G,X]= [G, X, X]$; (2) if either $X$ or $[G, X]$ is solvable, $G= [G,X]C_G(X)$;
(3) if $[G, X] \subseteq \Phi(G)$ then
$[G, X] = 1$.
\begin{quote}
\emph{Proof:}
\\
\emph{Claim:} $[g, x^i] = [g, x]^i \jmod{[G, X, X]}$.
\\
\emph{Proof of claim:} $[g, x^{i+1}] =
[g, x] [g, x^i]^x = [g, x] [g, x^i]^x [g, x^i, x]$ and this proves the claim.
\\
In particular, $1 = [g,1] =
[g, x^{|X|}]=
[g, x]^{|X|} \jmod{[G, X, X]}$ so $[G,X]/[G,X,X] =1$ which proves (1).
\\
\\
For (2), form the semi-direct product, $GX$.  $[G,X] \lhd GX$.  Put $G_0 = [G, X]X$.
\\
\emph{Claim:} $G_0 \lhd G$.
\\
\emph{Proof of claim:}
Suffices to show $[G, G_0] \subseteq G_0$.  Let $g \in G, g_0 \in G_0$ and put
$g_0 = g_1 x$ then $[g, g_0] [g, g_1]^x \in G_0$.  $G_0$ satisfies the Schur-Zassenhaus theorem.
Let $g \in G$ then $X^g \subseteq G_0^g = G_0$ and $X^g$ and $X$ are complements so
$X^{g g_1} = X, g_1 \in [G, X]$ and $[X, g g_1] \in  X \cap G = 1$.
$g g_1 \in C_G(X)$ and $g= c g_1^{-1}, c \in C_G(X) \in C_G(X)[G,X]$.
\\
\\
For (3), $G= [G,X] C_G(X)$.  If $\Phi(G) \supseteq [G,X]$,
$G=
\langle [G, X], C_G(X) \rangle \subseteq
\langle \Phi(G), C_G(X) \rangle =  C_G(X)$, so $[G, X]= 1$.
\end{quote}
{\bf Theorem 20:}
If $\phi$ is an irreducible representation of an abelian group $G$ with kernel $K$, then
$G/K$ is cyclic.
\begin{quote}
\emph{Proof:}
\emph{Claim:} If $G$ is abelian and acts irreducibly on $V$ then $gv= \lambda v$.\\
\emph{Proof of claim:}  $W= \{ w \in V: gw= \lambda w \} \ne 0$ is $G$-invariant,
so it is all of $V$.  This proves the claim.\\
Now
$G/K$ acts faithfully and irreducibly its letting $\lambda$ be the
$|G|$-th root of unity in the claim, $g \mapsto \Lambda(g)$.  This is an injection into
a subgroup of a multiplicative subgroup of $F$ and these subgroups are cyclic.
\end{quote}
{\bf Corollary:}  If a $G$ possesses a faithful irreducible representation, then
${\mathbb Z}(G)$ is cyclic.
\begin{quote}
\emph{Proof:}
Simple application of previous theorem.
\end{quote}
{\bf Lemma:} Let $P$ be an elementary abelian $p$-group and $Q$ be a non-cyclic
abelian $q$-group of $Aut(P)$, $p \ne q$, then $P= \prod_{x \in Q^{\#}} C_P(x)$.
\begin{quote}
\emph{Proof:}
Regard $P$ as a vector space over $F_p$.  $Q$ acts on $P$ as a linear transformation
and denote $\varphi: x \mapsto \phi_x$ be the map between $Q$ and the linear transformations
under this correspondence.
By Maschke, $P= P_1 \oplus \ldots \oplus P_n$, each $P_i$ irreducible.  Since
$Q$ is abelian and let $Q_i= ker(\varphi_{|P_i})$ under the
restricted map $\varphi_{|P_i}: Q \rightarrow P_i$.  $Q/Q_i$ must be cyclic and
$Q$ is non-cyclic so $Q_i \ne 1$.  If $x \in Q_i$ then $P_i \subseteq C_P(x)$.
\end{quote}
{\bf Theorem 21:}
Let $G$ be a $p$-group and
$f: Aut(G) \rightarrow Aut(G/\Phi(G))$.
$ker(f)= \langle \{ X: [G, X] \subseteq \Phi(G) \} \rangle$ and so $ker(f) = 1$.
\begin{quote}
\emph{Proof:}
This is practically the definition.
\end{quote}
{\bf Theorem 22:}
If $X$ acts on $G$, $(|X|, |G|)= 1$ and $G_0 \lhd \lhd G$, $C_G(G_0) \subseteq G_0$ and
$[G_0 , X] =1$ then $X$ acts trivially on $G$.
\begin{quote}
\emph{Proof:}
Suppose $G_0 \lhd G_1 \lhd \ldots \lhd G_n = G$.  Choose $i$ maximal such that
$[G_i, X]= 1$ (we want $i=n$).  $N_G(G_i)$ is $X$-invariant.  If
$N_G(G_i) \nsubseteq G$, we get $[N_G(G_i ), X]= 1$; but $G_{i+1} \subseteq N_G(G_i )$ which
implies $[G_i, X]= 1$ --- a contradiction.
$C_G(G_i ) \subseteq C_G(G_0) \subseteq G_0 \subseteq G_i$.
So $[G_i, X, G]=1= [G, G_i, X]$ and by the three subgroups lemma,
$[X,G,G_i]= 1$ and $[X,G] \subseteq C_G(G_i) \subseteq G_i$.
Therefore, $[G,X,X]=1$ and hence $[G,X]= 1$.
\end{quote}
{\bf Definitions:} $G$ acts \emph{irreducibly} on $V$  if there are no proper $G$-invariant
subgroups of $V$.
$G$ acts \emph{faithfully} on $V$ if $C_G(V) = 1$.
$G$ acts \emph{non-trivially} on $V$ if $C_G(V) \ne G$.
For $\alpha \in Aut(G)$, define $\sigma(G)= \sum_{g \in G} \alpha(g)$.
\\
\\
{\bf Theorem 23:}
If $G$ has a faithful, irreducible representation on a vector space $V$ over $F$,
$char(F) = p$ then $G$ has no non-trivial normal $p$-subgroup.
\begin{quote}
\emph{Proof:}
Let $\varphi$ be such a representation and $P \lhd G$.  $\varphi(P)$ fixes some
$v \ne 0$.  Set $W= C_V(\varphi(P))$.  $W$ is $G$-invariant and by irreducibility,
$W=G$ and $\varphi$ acts trivially on $G$ but $\varphi$ is faithful so
$P= 1$.
\end{quote}
{\bf Theorem 24:}
Let $G$ act on $V$, $V$, abelian with $(|G|, |V|)= 1$ then $C = V_1 \oplus V_2$ is the
direct sum of $G$-invariant subgroups of $V$ with $V_1= C_V(G)$.
\begin{quote}
\emph{Proof:}
Let $\sigma= \sigma(G)$.  Set $V_1 = C_V(\sigma)$ and $V_2= im(\sigma)$. $V_1$ and
$V_2$ are $G$-invariant and $|V_1| \cdot |V_2| = |V|$.  Set $Z= C_V(G)$, $Z$ is $G$-invariant
and $Z \subseteq V_2$.   \emph{Claim:} $Z \cap V_1 = 0$.  \emph{Proof of claim:}
Let $v \in Z \cap V_1$ then $\sigma(v)= |G|v$ and $\sigma(v)= 0$ hence $v= 0$.
\end{quote}
{\bf Theorem 25:}
Let $G$ act non-trivially on $V$, abelian, $(|G|, |V|)=1$, then $\exists W \subseteq V$,
$G$-invariant, on which $G$ acts non-trivially and irreducibly.
\begin{quote}
\emph{Proof:}
$V= [G,V] \oplus C_V(G)$ and $[G,V] \ne 1$.  By Maschke, $[G,V]$ decomposes into
irreducible modules.
\end{quote}
{\bf Theorem 26:}
If $A$ is a $p'$-group of automorphisms of the abelian $p$-group $P$ which
acts trivially on $\Omega_1(P)$ then $A= 1$.
\begin{quote}
\emph{Proof:}
$P= C \times H$ where $C= C_P(A)$ and $H= [P, A]$.  By hypothesis,
$\Omega_1(P) \subseteq C$ so $\Omega_1(P) \subseteq \Omega_1(C)$ and
$\Omega_1(P) = \Omega_1(C) \times \Omega_1(H)$ so $\Omega_1(H) = 1$.
So $H = 1$ and $P= C$  thus $A$ acts trivially on $P$ and $A \subseteq Aut(P)$.
and $P = 1$
\end{quote}
{\bf Theorem 27:}
If $A$ is a $p'$-subgroup of $Aut(P)$ and $1 \ne \phi A$, $\phi$ acts trivially
on every proper $A$-invariant normal subgroup of $P$, then
(1) $P' \subset {\mathbb Z}(P)$;
(2) $P/P'$ is an elementary abelian subgroup and $A$ acts irreducibly on $P/P'$;
(3) Either $P$ is elementary abelian or $P$ has class $2$, $P' = {\mathbb Z}(P) = \Phi(P)$
is elementary abelian and $\phi$ acts trivially on $P'$.
\begin{quote}
\emph{Proof:}
First, $\phi$ acts trivially on $P'$ since it is a proper $A$-invariant normal
subgroup of $P$.  $\phi$ does not act trivially on $P/P'$ because if it were,
$P \supset P' \supseteq 1$ of $P$ and $\phi= 1$ which it is not.
Suppose ${\overline P}= P/P' = {\overline P}_1 \times {\overline P}_2$ where
${\overline P}_1 \ne 1$ and
${\overline P}_i$ is $A$-invariant.  Then $P= P_1 P_2$ and $P_i$ is a proper
$A$-invariant normal subgroup of $P$ where $P_i$ is the inverse image of ${\overline P}_i$.
If $\phi$ is trivial on $P_1$ and $P_2$ then $\phi$ is trivial on $P$.
Thus we can assume $A$ acts indecomposiblyon ${\overline P}$.
If $\Omega_1({\overline P}) \subset {\overline P}$ then $\phi$ acts trivially on the
inverse image of ${\overline P}$ and $A$ acts trivially on
$\Omega_1({\overline P})$ but then $\phi$ acts trivially on ${\overline P}$.
Thus ${\overline P} = \Omega_1({\overline P})$.  From Maschke, $A$ acts irreducibly on
${\overline P}$.  Now, put $B= \langle \phi^A \rangle$.  Since $\phi$ acts trivially
on $P'$, $B$ acts trivially on $P'$.  Set $H= [P, B]$.  If $H \subseteq P'$, we have
$1 = [H,B]= [H, B, B]$ and so $B=1$.  Thus $H \nsubseteq P'$.
Moreover, since $B$ and $P$ are $A$-invariant and $H \lhd P$ and thus $H$ is $A$-invariant.
If $H \subset P$ then $\phi$ acts trivially on $H$ and hence on
${\overline H}$ but ${\overline H}$ is a non-trivially of ${\overline P}$
hence ${\overline H} = {\overline P}$ by irreducible action of $A$ on ${\overline P}$ and
thus $H = P$.  Since $B$ centralizes $P'$ and $P' \lhd P$, $[P',P,B] = 1 = [P, P', B]$
so $[P, B, P'] = 1$.  Since $[P, B] = P$ and $P'$ centralizes $P$
and $P' \subseteq {\mathbb Z}(P)$ which proves 1.
Assume $P$ is elementary abelian.  But $P' \ne 1$ and so $P' \subseteq {\mathbb Z}(P) \subset P$.
Now if $Z= {\mathbb Z}(P)$, ${\overline Z}$ is a proper $A$-invariant subgroup.  So
${\overline Z} = 1$ by the $A$-irreducibility.  Hence ${\mathbb Z}(P) = P'$ and
$cl(P)=2$.  Since $P' \subseteq \Phi(P)$, $P' = \Phi(P)$.
If $x, y \in P$ then $[x, y]= z \in  P'= {\mathbb Z}(P)$  so $[x,y^p]= z^p$.  But
$y^p \in {\mathbb Z}(P)$ and then $z^p= 1$ and $[x,y]^p= 1$.  Since $P'$ is abelian,
it follows that $P'$ is elementary abelian and the proof is complete.
\end{quote}
{\bf Theorem 28:}
Let $G$ be a $p'$-group acting non-trivially on an abelian $p$-group $V$ then $G$ acts
non-trivially on $V_0 \subseteq V$ with $V_0 = \{ v: pv= 0 \} = \Omega_1(V)$.
\begin{quote}
\emph{Proof:}
By induction on $|P|$.  If $Q \le P, \Omega_1(Q) \le \Omega_1(P)$.  So if
$Q$ is any proper $A$-invariant subgroup, the action is trivial on $\Omega_1(Q)$
and $P$ is a special $p$-group (see chapter 15) and $A$ acts trivially on $P'$ and
irreducibly on $P/P'$ so $cl(P) \le 2$.  Hence,
$x \in P$ and $\varphi \in  A$, implies $ (\varphi(x)x^{-1})^p = (\varphi(x))^px^{-1)^p}) $
but then $x^p \in P'$ and $\varphi$ fixes $x^p$.  Since $[P,A] = \langle \varphi(x) x^{-1} \rangle$,
$[P, A] \subseteq \Omega_1(P)$ and $A$ would stabilize the series
$P \supseteq \Omega_1(P) \supseteq 1$ and thus $A=1$.
\end{quote}
{\bf Theorem 29:}
Let $G$ act faithfully and irreducibly on an abelian group of $V$, $(|G|, |V|)= 1$ then
${\mathbb Z}(G)$ is cyclic.
\begin{quote}
\emph{Proof:}
If ${\mathbb Z}(G)$ is not cyclic it contains a subgroup $H$ of type $(p, p)$ hence
by an earlier result, $H= \bigcup_{i} H_1, H_i \cap H_j = \{ 0 \}$ since
$p \nmid |V|$ then $pV \ne \{ 0 \}$, $\exists x_i: C_V(x_i ) \ne 0$.  Now $G$
acts faithfully so $C_V(x_i) \ne V$ and $C_V(x_i)$ is $G$-invariant.  This is a contradiction.
\end{quote}
{\bf Theorem 30:}
Let $P$ be a $p$-group and $Q$ be a non-cyclic abelian $q$-group of automorphisms acting on a $P$,
$q \ne p$
then $P= \prod_{x \in Q^{\#}} C_P(x)$.
\begin{quote}
\emph{Proof:}
By induction on $|P|$.
Let $x_j$ be a fixed ordering of $Q^{\#}$.
Put $Z= \Omega_1({\mathbb Z}(P))$
and $Z_j= C_Z(x_j)$, some $j$,
then $Z \ne 1$ by the previous result on abelian $P$.
The theorem holds by induction on ${\overline P}= P/Z_j$ and
${\overline {C_P(x_i)}}= C_{\overline P}(x_i)$,
so $P= Z_j \prod_{i=1}^n C_P(x_i)$ but
$Z_j \subseteq C_P(x_j) \subseteq P$.  This proves the result.
\end{quote}
{\bf Theorem 31:}
Let $\varphi$ be a representation of $G$ on a vector space $V$ with $char(F)=0$ or
$(char(F), |G|) = 1$. If
$ V= V_1 \supset V_2 \supset \ldots \supset V_n \supset V_{n+1}= 0$ a sequence of $G$-invariant
subspaces and $\varphi(G)$ acts trivially on each $V_i/V_{i+1}$ then $\varphi$ is the trivial
representation.
\begin{quote}
\emph{Proof:}
$V= V_n \oplus W$ and $W$ is $G$-invariant.  $\varphi_{|W}$ is equivalent to the
quotient representation, ${\overline {\varphi}}$ of $G$ on $V/V_n$ but then
induction on $dim_F(V)$ yields $\varphi_{W}$ is trivial and since $\varphi_{V_n}$ is also
trivial, $\varphi$ is trivial.
\end{quote}
{\bf Theorem 32:}
Let $A$ be a regular group of automorphisms of a $p$-group $P$ then
(1) $A$ is a $p'$-group;
(2) $A$ possesses no non-cyclic abelian subgroups;
(3) A subgroup of $A$ of order $qr$ is cyclic.
\begin{quote}
\emph{Proof:}
Let $B \in S_p(A)$.  $B$ acts on $V= \Omega_1({\mathbb Z}(P))$ which is elementary abelian
and so $C_V(B) \ne 1$.  Since elements of $A^{\#}$ only fix $1 \in G$, this forces
$B= 1$ and $A$ is a $p'$-group. (2) follows from the fact that
$V= \langle C_V( \varphi): \varphi \in Q^{\#} \rangle$ and this forces $C_V(\varphi) \ne 1$
contradicting regularity.  For (3), let $D$ be such a subgroup. $q \ne r$ is obvious so
assume $q > r$ and $Q \in S_q(D), R \in S_r(D)$.  $D= QR$ and $Q \lhd D$.  If
$C_D(Q) \supset Q$ then $C_Q(D)= D$ and $D$ is abelian and hence cyclic.  Hence we
can also assume $C_D(Q)= Q$.  Applying Thompson's $p \times q$ theorem to the
action of $D$ on $V$ gives $C_V(R) \ne 1$ which again contradicts regularity.
\end{quote}
\section {$A$-invariance}
{\bf Lemma:}
If $G$ is a group acting on a group $V$ and $H \lhd G$ then $C_V(H)$ is $G$-invariant.
\begin{quote}
\emph{Proof:}
Put $W= C_V(H)$.  For $w \in W, h \in H$ and any $x \in G$, $w^{xhx^{-1}} = w$ since
$xhx^{-1} \in H$.  Thus  $(w^x)^h = (w^x)$ and $w^x$ is fixed by $h$, that is $w^x \in W$.
\end{quote}
{\bf Theorem 33:}
Let $G$ be a $p$-group acting on an elementary abelian $p$-group $V$ (Hence $G$ is a set of
invertible transformations in $GL_n(p)$.)  Then $\exists v \in V, v \ne 0: v \in C_V(G)$.
\begin{quote}
\emph{Proof:}
By induction on $|G|$.  Let $M$ be a maximal subgroup of $G$.  $|G:M|= p$ and $M \lhd G$.
Put $W= C_V(M)$.  By induction, $W > \{1 \}$ and by the lemma, $W$ is $G$-invariant.
Choose $y \in G \setminus M$.  The minimal polynomial for $y$, $min(y) \mid (x^p-1)$ since
$y^p \in M$.  So $\exists w_1 \in W, w_1 \ne 0: w_1^y = w_1$ and we already know
$w_1^m = w_1, m \in M$ so $w_1 \in C_V( \langle y, M \rangle)= C_V(G)$ and we're done.
\end{quote}
{\bf Note:} In this section, $A$ acts on $G$ and $(|A|, |G|)=1$ with either
$A$ or $G$ solvable.  Note that $A/C_A(G)$ acts faithfully on $G$.
\\
\\
{\bf Theorem 34:}
Suppose that $A$ is an elementary abelian $p$-group such that $r(A) \ge 3$.  If $P, Q$ are
$A$-invariant $p'$-groups, $\exists a \in A$ such that
$C_P(a) \ne 1$ and $C_Q(a) \ne 1$.
\begin{quote}
\emph{Proof:}
Let $V$ be a subgroup of type $(p, p)$.  Since
$P= \langle C_P(v): v \in V^{\#} \rangle$, $\exists v \in V: C_P(v) \ne 1$.  Let
$W \subseteq A$ such that $W$ is of type $(p, p)$ and $W \cap \langle v \rangle = 1$.
$\exists w \in W: C_P(w) \cap C_P(v) \ne 1$ since $C_P(v)$ is $W$-invariant.
Then $\langle v, w \rangle$ is of type $(p, p)$ and acts on $Q$.
Thus $\exists a \in \langle v, w \rangle^{\#}$ such that  $C_Q(a) \ne 1$.
Since $C_P(a) \supseteq C_P(w) \cap C_P(v) \ne 1$, we are done.
\end{quote}
{\bf Theorem 35:}
If $U \leq G$ is $A-$invariant and $g$ satisfies $(Ug)^A=Ug$ then
$\exists c \in C_G(A)$: $Ug=Uc$. If $N$ is an $A-$invariant normal subgroup of $G$ then
(1) $C_{G/N}(A)= C_G(A)N/N$ (This shows $G=[G,A]C_G(A)$.) and (2) if $A$ acts
trivially on $N$ and $G/N$ then $G$ acts trivially on $G$. If $p \mid |G|$ (the analogous
results hold for $\pi$) then (1) $\exists S \in S_p(G): S^A=S$, (2) all such $A-$invariant
Sylow $p-$groups are conjugate under $C_G(A)$, (3) every $A-$invariant $p$-group of
$G$ is contained in an $A-$invariant Sylow $p-$group.
\begin{quote}
\emph{Proof:}
Suppose $g \in G$ then $(UG)^A = Ug$ so $[g,a] \in U, \forall a \in A$.
$A^{g^{-1}}a \subseteq AU$ and $A$ and $A^{g^{-1}}$ are complements $U$ in $AU$
so $\exists n \in U: A= A^{gn}$ and the result holds.
\end{quote}
{\bf Theorem 36:}
Let $N$ be an $A$-invariant normal subgroup of $G$, $(|A|, |G|)=1$ then
(a) $C_{G/N}(A)= C_G(A)N/N$ and  (b) if $A$ acts trivially on $N$ and $G/N$ then
$A$ acts trivially on $G$.
\begin{quote}
\emph{Proof:}
By the previous result, $\exists c \in C_G(A): Ng= Nc$.  This proves (a).(b) follows from (a).
\end{quote}
{\bf Theorem 37:}
Let $N$ be an $A$-invariant normal subgroup of $G$, $(|A|, |G|)=1$ then
if $A$ acts trivially on $N$, it acts trivially on $G/C_G(A)$.
\begin{quote}
\emph{Proof:}
Since $[N,A]=1$, $[N,A,G]= 1= [G,N,A]$ and so $[A,G,N] = 1$.
\end{quote}
{\bf Theorem 38:}
Let $G$ act on $\Omega$ and $K \lhd G$ with (1) $(|K|, |G/K|)=1$, (2) $K$ or $G/K$ solvable,
(3) $K$ acts transitively on $\Omega$.  Then for evey complement, $H$ of $K$, (1) $C_{\Omega}(H) \ne \emptyset$ and
(2) $C_K(H)$ acts transitively on $C_{\Omega}(H)$.
\begin{quote}
\emph{Proof:}
For (a), let $\beta \in \Omega$.  Since $K$ is transitive, $|\Omega| \mid |K|$ and $G= K G_{\beta}$.
$G/K \approx G_{\beta}/(G_{\beta} \cap K)$.  Apply Schur-Zassenhaus to $G_{\beta} \cap K$ to get a complment
$H_1$ with $\beta \in C_{\Omega}(H_1)$, $H_1(G_{\beta} \cap K)= G_{\beta}$ and $H_1K=G$.  Since all complements
conjugate to $H_1$, (a) is established.  For (b), let $\alpha, \beta \in C_{\Omega}(H), k \in K$ with
$\alpha^k= \beta$.  $H$ and $H^k$ are two complements of $K \cap G_{\beta}$ in $G_{\beta}$ so
$\exists k' \in G_{\beta}$ such that $\alpha^{k k'}= \beta$.  $[k k', H] \leq H \cap K =1$ so
$k k' \in C_K(H)$.
\end{quote}
{\bf Theorem 39:}
Let $p \mid |G|$ and suppose the action of $A$ on $G$ is co-prime.  (1) There is an $A$-invariant Sylow $p$-group of
$G$. (2) The $A$-invariant Sylow $p$-groups are conjugate under $C_G(A)$.  (3) Every $A$-invariant $p$-subgroup is
contained in an $A$-invariant Sylow $p$-subgroup of $G$.
\begin{quote}
\emph{Proof:}
The semi-direct product $AG$ acts on $S_p(G)$ and (1) and (2) follow from previous result.
For (3), let $U$ be a maximal $A$-invariant $p$-subgroup of $G$, we show $U \in S_p(G)$.
Assume not.  Then $U \notin S_p(G_1)$, $G_1 = N_G(U)$.  $G_1$ is $A$-invariant and $\exists T \in S_p(G_1)$
by (1).  $U<T$ which contradicts maximality.
\end{quote}
{\bf Theorem 40:}
If $T=\bigcap_{S \in S_p(G), S^A=S} S$,
then $T$ is the largest $A-$invariant $p-$subgroup of $G$ normalized by $C_G(A)$.
\begin{quote}
\emph{Proof:}
By previous result, $\exists S \in S_p(G): S^A=A$ and
$\{P \in S_p(G): P^A=P \} = \{S^c: c \in C_G(A) \}$, so the intersection of theh Sylow groups is
$A$-invariant. By previous result, any $A$-invariant $p$-subgroup is contained in an $A$-invariant
Sylow $p$-group of $G$.  If $U$ is normalized by $C_G(A)$, then $U$ is contained in every $A$-invariant
Sylow $p$ group and hence their intersection.
\end{quote}
{\bf Theorem 41:}
If $P$ is an $A-$invariant Sylow $p-$group of $G$ and $H \le G$ with $H^A=H, H^{C_P(G)}=H$ then
$P \cap H \in S_p(H)$.  If $A= P \times Q$ acts on $M$ and $P, M$ are $p-$groups and
$Q$ is a $p'-$group with $C_M(P) \le C_M(Q)$ then $[M,Q]=1$. If $A$ acts trivially on
$G/\Phi(G)$ then $A$ acts trivially on $G$ and if $\Phi(G)$ is a $p-$group then
so is $A/C_A(G)$.
\begin{quote}
\emph{Proof:}
$G= \Phi(G) C_G(A)$ so $G= C_G(A)$.
\end{quote}
{\bf Applying $P \times Q$:} If $p \in \pi(G)$ and ${\overline G}= G/O_{p'}(G)$
with $C_{\overline G}(O_p({\overline G})) \le O_p({\overline G})$ then
$\forall P \in p(G), O_{p'}(N_G(P))=O_{p'}(G) \cap N_G(P)$.
\\
\\
{\bf Thompson:} Let $a$ be a $p'$ automorphism of a $p$ group $G$ and suppose
$X$ is a $p$-group of automorphisms of $G$ and
$[a,X]=1=[a,C_G(X)]$ then $a=1$.
\begin{quote}
\emph{Proof:}
Let $N \subseteq G$ be and $X$-invariant subgroup such that $[a,N] \ne 1$ but
$[a, K]= 1$ for all $X$-invariant proper subgroups $K$ of $N$.  Since
$[N,X,a]=1$ and $[X,a,N]=1$,  $[a, N, X] = 1$, $[N,a] \subseteq C_G(X)$,
$[N,a,a]= 1$ and thus $[N,a]= 1$.
\end{quote}
{\bf Theorem 42:} Let $a$ be a $\pi'$ automorphism of a $\pi$ group $G$ and suppose
$X \lhd \lhd G$ such that $[a,X]=1=[a,C_G(X)]$ then $a=1$.
\begin{quote}
\emph{Proof:}
Let $X \lhd X_1 \lhd \ldots \lhd X_n=G$ and choose $i$ so that
$[a, X_{i+1}] \ne 1$ but
$[a, X_{i}] = 1$.  Let $N= N_G(X_i )$.  Since $X_{i+1} \subseteq N$, $[a,N] \ne 1$, yet
$[X_i, N, a] =1$ and
$[X_i, a, N] =1$ so
$[a, N, X_i] =1$ and $[N,a] \subseteq C_G(X_i ) \subseteq C_G(X)$.  Hence
$[N,a,a] = 1$ and $[N,a] = 1$.
\end{quote}
{\bf Theorem 43:}
If $P$ is a $p-$group and $Q$ a $p'-$ group with $Q \mapsto Aut(P)$ then
$Q$ is faithful on $P/\Phi(P)$.
\begin{quote}
\emph{Proof:}
Suppose $b \in A$ centralizes $G/\Phi(G)$.  If $b \ne 1$, there is a non-trivial
power of $b$ which is a $q$-element and which centralizes $G/ \Phi(G)$.  Let
$B= \langle b \rangle $ and $g \in G$.  $B$ acts on the coset $Xg$.  $|Fix_B(Xg)| = |X| \jmod{q}$.
$X= \Phi(G)$ has order that is a power of $p \ne 0 \jmod{q}$, so $b$ centralizes some
element $x \in X$.  So $G= \langle Y \rangle \le C_G(B)$ where
$Y$ is a set of coset representatives,
thus $B=1$.
\end{quote}
{\bf Theorem 44:}
If $A$ acts on $G$, $(|A|,|G|)= 1$ and $G$ be abelian, $A$ is faithful on $\Omega_1(G)$.
\begin{quote}
\emph{Proof:}
WLOG, $[A, \Omega_1(G)]=1$.  Let $X$ have order $p$, ${\overline G}= G/X$.  $A$ is
faithful on $\Omega_1(G)$ so $A$ is faithful on $\Omega_1({\overline G})$ so WLOG
${\overline G}= \Omega_1({\overline G})$.  By maximality, $C_{\overline G}(A)=1$ so
$X=C_G(A)$, $X= \Omega_1(G)$ and so $G$ is cyclic.  This is not possible given the
automorphism group of a cyclic group.
\end{quote}
\section {More Results}
{\bf Definition:} $G$ is $p$-solvable if $G$ has a normal series whose quotients consist of $p$-groups
and $p'$-groups.
\\
\\
{\bf Definition:} $G$ is an $A_p$-group if $\forall S, T \in S_p(G)$, $\exists x \in C(S \cap T)$ such that
$S^x = T$.  For $X \in p(G)$, define $A(X) = N_G(X)/C_G(X)$.
\\
\\
{\bf Theorem 45:} If $G$ is a solvable group and $H$ is a minimal normal subgroup then $H$ is elementary abelian.
\begin{quote}
\emph{Proof:}
$H$ is characteristically simple so $H' = 1$ and $\Omega_1(H) = H$ so $H$ is elementary abelian.
\end{quote}
{\bf Theorem 46:} If $\alpha \in Aut(G)$ and $\alpha$ acts trivially on $G/\Phi(G)$ then
$G$ acts trivially on $G$.
\begin{quote}
\emph{Proof:}
Suffices to assume that $\alpha$ has order $q \neq p$.  Let $x_1 , x_2 , \ldots x_r$ be a minimal generating set.
Since $\alpha$ fixes each coset, $\alpha$ acts on each coset.  Since $q \neq p$, $\alpha$ must fix an element
of each coset. Say $y_i$ is fixed in $\Phi(G) x_i$.  $\langle y_1, \ldots y_r \rangle = G$ and the result holds.
\end{quote}
{\bf 1.2.3 Theorem:} If $G$ is $p$-solvable with $O_{p'}(G) = 1$ then $C_G(O_p(G)) \subseteq O_p(G)$.
\begin{quote}
\emph{Proof:}
Put $H = O_p(G) C_G(O_p(G)) \lhd G$.  If $O_{p'}(H) = 1$: if not, $O_{p'}(H) \subseteq O_{p'}(G) = 1$
which is a contradiction.  Suppose $H > O_p(G)$.  $H/O_p(G)$ is $p$-solvable.  $H/O_p(G)$ cannot have
a normal $p$-group since it would be in $O_p(G)$.  Let $K$ be the inverse image of $O_{p'}(H/O_p(G))$.
$K \lhd G$ and $K > O_p(G)$.  $O_p(G) \in S_p(K)$ so $\exists L \in p'(G)$: $K = LO_p(G)$.  Let $x \in L$,
$x = yz, y \in O_p(G), z \in C(O_p(G))$.  $[x, O_p(G)] = [y, O_p(G)] =1$ so $K \subseteq O_{p'}(G)$, contradiction.
\end{quote}
{\bf Theorem 47:} If $N \subseteq G$ has a normal $p$-complement ($G = P N, N \lhd G$) then $G$ is an
$A_p$ group.  If $X \in p(G)$ then $A(X)$ is a $p$-group.
\begin{quote}
\emph{Proof:}
Suppose $S, T \in S_p(G)$, $G = SN = TN$.  $\exists y \in N: S^x = T$.  Suppose $s \in S \cap T$, $S^y \in T$
so $[s, y] \in T$,  so $[s, y] = 1$ and $y \in C(S \cap T)$.  Let $X \in p(G)$.  $M = N_G(X)$, then $M \cap N$
is a normal $p$-complement of $M$.  If $x \in X$ and $y \in M \cap N$ then $[x,y] \in X \cap (M \cap N)$, so
$y \in C(X)$ and $M \cap N \subseteq C(X)$.  Thus $y$'s action is as a $p$-group and $A(X)$ is a $p$-group.
\end{quote}
{\bf Frobenius Theorem:} $G$ be group. If $\forall X \in p(G)$, $A_G(X)$ is a $p$-group then $G$ has a normal
$p$-complement.
\begin{quote}
\emph{Proof:}
Let $G$ be a minimal counterexample.
\\
\\
1. If $H < G$ then $H$ has a normal $p$-complement.
\\
Let $X \in p(H)$, $A_H(X) = N_H(X)/C_H(X) \rightarrow A_G(X)$, so $H$ satisfies the hypothesis of theorem and
$H$ has a normal $p$-complement.
\\
\\
2. $O_p(G) = 1$.
\\
Let $K = O_p(G)$.  $X/K \in p(G/K)$.  So $A_G(X) \rightarrow A_{G/K}(X/K)$.  $G/K$ has a normal $p$-complement
$(M/K) (X/K) = G/K$.  $M \lhd G$ and $K \in S_p(M)$, $K \lhd M$, $LK = M$ and $[L,K] \subseteq K$.
$M = L \times K$,
$L = O_{p'}(M)$,  $M \lhd G$ and $L \lhd G$.  $L$ is a normal $p$-complement.  Contradiction.
\\
\\
3. $G$ is an $A_p$-group
\\
Let $P, P_1 \in S_p(G)$.  Proof is by induction on $|P: P \cap P_1|$.  $P \neq P_1$ and $P \cap P_1 \neq 1$.
Put $H = N_G(P \cap P_1 )$.  Since $O_p(G) = 1$, $H \neq G$ and $H$ has a normal $p$-complement.  If $P > Q$.
$N_P(Q) > Q$.  Pick $R, R_1 \in S_p(H)$ then $H \cap P \subseteq R$ and
$H \cap P_1 \subseteq R_1$.  $\exists y \in C_H(R \cap R_1): R_1 = R^y$.  Let $S \in S_p(G)$ with
$R \subseteq S$.  Set $S_1 = S^y$ so
$R_1 \subseteq S_1$.   $P \cap S > P \cap P_1$ and $P_1 \cap S_1 > P \cap P_1$
By induction,
$\exists u \in C_G(S \cap P),
\exists v \in C_G(S_1 \cap P_1):$ $P^u=S$ and $S_1^v=P_1$ .  $P^{uyv} = P_1$ and $uyv \in C(P \cap P_1)$.
Since $R \cap R_1$, $S \cap P$ and $S_1 \cap P_1$ all contain $P \cap P_1$, $uyv \in C_G(P \cap P_1 )$.
\\
\\
4. Contradiction
\\
Let $P \in S_p(G), g,h \in P$, $h= g^x$ and $h \in P \cap P_1$.  Since $G$ is a $A$-group,
$\exists y \in C(P \cap P^x)$ with $P^{xy}=P$, $h^y=h$.  Now $xy \in N(P)= C(P)P$.
Since $G$ is an $A_G(P)$ is a $p$-group, $xy=uv$, wtih $u \in C(P)$ and $ v \in P$.
$g^v = g^{uv} = g^{xy} = h^y = h$.  Thus any two elements conjugate in $G$ are conjugate in $P$ so $G$ has a normal
$p$-complement.
\end{quote}
{\bf Theorem 48:} Let $G$ be a group with no normal $2$-complement and $Q \in S_2(G)$.
Suppose $Q$ has a cyclic subgroup,
$A$ of index $2$.  Then $ 3 \mid |G|$.
\begin{quote}
\emph{Proof:}
Since $G$ does not have a normal $p$-complement, $\exists X \subseteq Q$ such that $N(X)/C(X)$ is not a
$2$-group. $X \cap A$ is cyclic and $|X: A \cap X|$ is 1 or 2 and $X$ has two generators.
$\exists g \in N(X)/C(X)$ of order
$q, q \neq 2$ that acts faithfully on $X$ and hence $X/\Phi(X)$.  Since $X$ has 2 generators, the Burnside
Basis Theorem yields $|X/\Phi(X)| \leq 4$ and $q = 3$.
\end{quote}
{\bf Definition:} Let $S$ be a  $p$-group.  For $A: A' = 1$, $m(A)$ is the minimal number of generators of $A$.
$d(P)$ is the maximum value of $m(A)$ for $A \subseteq S$,
$A' = 1$. $J(S) = \langle A: m(A) = d(S) \rangle$.
${\cal H}(G)$ be a set of $p$-groups, $H \subseteq G$: $N_G(H)$ does not have a normal $p$-complement.
Thompson ordering: For $H, K \in G$, define $H \leq \leq K$ iff
(a) $|N_G(H)|_p \leq \leq |N_G(K)|_p$, or
(b) $|N_G(H)|_p = |N_G(K)|_p$ and $H < K$, or
(c) $H = K$.
\\
\\
{\bf Theorem 49:}   Let $G$ be a $p'$ group acting non-trivially on an abelian $p$-group, $V$ with
$(|G|, |V|) =1$.  $\exists W \subseteq V$ on which $G$ acts non-trivially and irreducibly.
\begin{quote}
\emph{Proof:}
$V = V_1 \times V_2$, $V_1 = ker(\sigma)$, $V_2 = im(\sigma)$, where
$\sigma(v)= \sum_{g \in G} v^g$.
Let $V_1$ and $V_2 \neq 0$ be as above.  Let $W \neq 0$ be a minimal $G$-invariant
subgroup of $V_2$.  Since $W \cap C_G(G) = 0$, $G$ acts nontrivially on $W$.
\end{quote}
{\bf Corollary 1:}   Let $G$ be a $p'$ group acting non-trivially on an abelian $p$-group, $V$.
Then $G$ acts non-trivially on $V_0 \subseteq V: V_0 = \{ v \in V: vp = 0 \}$
\\
\\
{\bf Corollary 2:}   Let $G$ act faithfully and irreducibly on and abelian group, $V$.
If $(|G|, |V|) =1$, then ${\mathbb Z}(G)$ is cyclic.
\begin{quote}
\emph{Proof:}
If not, $G$  has a central subgroup $H$ which is abelian of type $(p, p)$, some $p \mid |G|$.
$H$ is a disjoint union of $p+1$ subgroups $J_0, \ldots, J_p$ of order $p$. Since $p \nmid |V|$,
$vp \neq 0$ and $\exists i: C_V(J_i) \neq 0$.  $G$ acts faithfully on $V$ so $C_V(J_i) \neq V$.
Since $J_i \lhd G$, $C_V(J_i)$ is a proper $G$ invariant subgroup of $V$ and $G$ does not act irreducibly,
contradiction.
\end{quote}
{\bf Theorem (Thompson):}  Let $p$ be an odd prime.  If $S \in S_p(G)$ and both $C_G({\mathbb Z}(S))$ and
$N_G(J(S))$ have normal $p$-complements, so does $G$.  There is another proof of this in the
stability section.
\begin{quote}
\emph{Proof:}
Let $G$ be a minimal counterexample of minimal order.
\\
\\
1. ${\cal H}(G) \neq \emptyset$. Let $H$ be maximal in the Thompson ordering of ${\cal H}(G)$.  Put $N = N_G(H)$.
$P \in S_p(N)$ with $H \subseteq P \subseteq S$.
\\
\\
2. $H \neq S$.
\\
If $H = S$ then $N \subseteq N_G(J(S))$ which has a normal $p$-complement.
So $N$ has a normal $p$-complement.
Contradiction.
\\
\\
3. ${\overline N} = N/H$ has a normal $p$-complement.
\\
$P \in S_p(N)$.  Since $H \neq S$, $N_S(H) > H$ and $P > H$.
Let ${\overline P}$ be the image of $P$ in ${\overline N}$ then ${\overline P} \in S_p({\overline N})$.
Suppose ${\overline N}$ does not have a normal $p$-complement.  Since $|{\overline N}| < |G|$, either
$C_{\overline N}({\mathbb Z}({\overline P}))$ or $N_{\overline N}(J({\overline P}))$ does not
have a normal $p$-complement.
Let $K$ be the complete inverse image of either ${\mathbb Z}({\overline P})$ or
$J({\overline P})$, which ever one fails.
$K$ is a $p$-group and $N_G(K)$ does not have a normal $p$-complement.
$P \subseteq N_G(K)$ so either $|N_G(K)|_p > |N_G(K)|_p$ or
$|N_G(K)|_p = |N_G(K)|_p$ and $|K| > |H|$.  So $K \in {\cal H}$, $K \geq \geq H$ but $K \neq H$, contradiction.
\\
\\
4. $N = G$.
\\
$N$ satisfies the hypothesis. $H \subseteq P \subseteq S$,
${\mathbb Z}(S) \subseteq N_G(H) = N$. $P {\mathbb Z}(S) \subseteq N \cap S$
and $P {\mathbb Z}(S) = P$.  ${\mathbb Z}(S) \subseteq P$ so ${\mathbb Z}(S) \subseteq {\mathbb Z}(P)$ and
$C_N({\mathbb Z}(P)) \subseteq C_G({\mathbb Z}(P)) \subseteq C_G({\mathbb Z}(P))$, so $C_N({\mathbb Z}(P))$ has
a normal $p$-complement and so does $C_N({\mathbb Z}(P))$.
If $P = S$, $N_N(J(P) \subseteq N_G(J(S))$ has a normal $p$-complement.
Now suppose $P < S$.  Then $N_S(P) > P$ so $|N_G(J(P)|_p > |N|_p$.
By maximality of $H$, $J(P) \notin {\cal H}$
and so $N_G(J(P)$ has a normal $p$-complement and so does $N_N(J(P))$.
If $N \neq G$, since $G$ is minimal,
$N$ has a normal $p$-complement, contradiction.  So $N = G$.
\\
\\
5. $O_{p'}(G) = 1$
\\
Put $L = O_{p'}(G)$.  We show ${\overline G} = G/L$ satisfies the assumptions of the theorem.
Let $K \subseteq S$.  We show ${\overline {N_G(K)}} = N_{\overline G}({\overline K})$.  Let
${\overline x} \in N_{\overline G}({\overline K})$.  Since $L \lhd KL$, $K \in S_p(KL)$ so
$\exists a \in K, b \in L: K^x = K^{ab}= K^b$.
Hence $xb^{-1} \in N_G(K)$ and ${\overline {xb^{-1}}}= {\overline x}$.
${\overline {N_G(J(S)})} = N_{\overline G}(J({\overline S}))$ so
${\overline {J(S)}} = J({\overline S})$
and it has a normal $p$-complement.
${\overline {C_G(K)}} = C_{\overline G}({\overline K})$.  Conversely, let
${\overline x} \in C_{\overline G}({\overline K}) \subseteq N_{\overline G}({\overline K})$,
so we can assume $x \in N_G(K)$. Since $[x, K] \subseteq  K \cap L = 1$, $x \in C_G(K)$.
${\overline {{\mathbb Z}(S)}} = {\mathbb Z}({\overline S})$ so
${\overline {C_G({\mathbb Z}(S)}} = C_{\overline G(}{\mathbb Z}({\overline S})$
and it has a normal $p$-complement.
If $|L| > 1$, $|{\overline G}| < |G|$ so ${\overline G}$ has a normal $p$-complement
and its complete inverse image is a normal $p$-complement of $G$.  Contradiction.
\\
\\
6. $H = O_p(G)$ and $G$ is $p$-solvable of $p$-length $2$.
\\
Set $K = O_p(G)$ so $H \subseteq K$.  Since $G = N_G(K)$has no normal $p$-complement,
$K \in {\cal H}$.  By the
maximality of $H$ in ${\cal H}$ and the fact that $G=N_G(K)= N_G(H)$, $H=K$.
By 3 and 4, $G/H$ has a normal $p$-complement, so $G$ is $p$-solvable of length at most 2.
\\
\\
7.  ${\overline G} = G/H = {\overline {SM}}$ then ${\overline M}$ has a normal $p$-complement and
${\overline M}$ contains no proper ${\overline S}$-invariant subgroup.
\\
Let ${\overline M} > {\overline {M_0}} > 1$ and suppose ${\overline {M_0}}$ is
$S$-invariant.  Let $M_0$ be the complete inverse image of ${\overline {M_0}}$.  Put $G_0 = S M_0$,
so $G > G_0$.
Since
$C_{\overline {G_0}}({\mathbb Z}({\overline P}))$ or
$N_{\overline {G_0}}(J({\overline P}))$ have normal $p$-complements,
so does $G_0 = S K_0$.  $K_0 \lhd G_0, S \cap K_0 = 1, [H, K_0] \subseteq K_0, [H, K_0] \subseteq H$.  Hence
$[H, K_0] = 1$ and $K_0$ centralizes $H$, this contradicts 5, 6, and 1.2.3.
\\
\\
8. ${\overline M} = {\overline Q}$ is an elementary abelian $q \neq p$ group.  ${\overline S}$
acts irreducibly on ${\overline Q}$ and $S$ is a maximal subgroup of $G$.
\\
Let $q \mid |{\overline M}|$.  ${\overline S}$ permutes $S_q$ subgroups of ${\overline M}$ and the
number of such subgroups divides $|{\overline M}|$.  Some such orbit has size $1$, say it contains
${\overline Q}$.  By step 7, ${\overline M} = {\overline Q}$.
${\overline Q}$ has no proper $G$-invariant
subgroup so ${\overline S}$ acts irreducibly on ${\overline Q}$.  Finally, $G > L >S$ and
${\overline Q} > {\overline Q} \cap {\overline L} > 1$ and ${\overline Q} \cap {\overline L}$ is
a proper ${\overline S}$ invariant subgroup of ${\overline Q}$, which is a contradiction.  So
$S$ is maximal in $G$.
\\
\\
9. $\exists A$, abelian: $A \subseteq S$ with $m(a) = d(S)$ and with $A \nsubseteq H$.  Let $A$ be
fixed such group of minimal order.  Let $A_0 = A \cap H$ then $A/A_0$ is elementary abelian.
\\
If $J(S) \subseteq H$, $J(S) \; char \; H \lhd G$ so $G = N(J(S))$
has a normal $p$-complement, contradiction.
So $J(S) \nsubseteq H$.  Since $J(S) = \langle A \rangle, S \subseteq S$,
$A' =1$ with $m(A)=d(p)$,
at least one such $A \nsubseteq H$.  Let $A$ be one such of minimal order.
Set $A_0 = A \cap H$.  $A_1/A_0 = \Omega_1(A/A_0)$,
$A_1'=1$.  So $A_1 \nsubseteq H$. Also, $m(A)= m(A_1)$.
By minimality, $A = A_1$ and $A/A_0$ is elementary abelian.
\\
\\
10. Let ${\overline A} = AH/H$ then ${\overline G} = {\overline A} {\overline Q}$ and $|{\overline A}| = p$.
\\
first, we show ${\overline G} = {\overline A} {\overline Q}$.
Since ${\overline G} = {\overline S} {\overline Q}$ and
${\overline S} \subseteq N({\overline Q})$.  This is failthful since $H = O_p(G)$ and hence a normal subgroup
of ${\overline G}$.  ${\overline A} = AH/H \cong A/A_0 \neq 1$.
\\
Now, ${\overline A}$ acts non-trivially and faithfully on some
${\overline {Q_1}} \subseteq {\overline Q}$ by theorem 49.
Let $G_1$ be the complete inverse image in ${\overline G}$ of ${\overline A} {\overline {Q_1}}$.
$P_1 \in S_p(G_1)$, $A \subseteq P_1$. $P_1 \subseteq S$.
Since $H \subseteq P_1$, and $C_G(H) \subseteq H$, by
1.2.3, ${\mathbb Z}(S) \subseteq C_G(H) \subseteq H \subseteq P_1$ and ${\mathbb Z}(S) \subseteq {\mathbb Z}(P_1)$.  Hence
$C_G({\mathbb Z}(P_1)) \subseteq C_G({\mathbb Z}(S))$ and the latter has a normal $p$-complement.
Since $A \subseteq P_1$ and $m(A) = d(S)$, $D(P_1)=d(S)$.
Thus $A \subseteq J(P_1)$.  Let $Q_2 \in S_q(N_{G_1}(J(P_1))$.
$[A, Q_2] \subseteq [J(P_1), Q_2] \subseteq J(P_1)$ and $[A, Q_2]$ is a $p$-group.
${\overline {Q_2}} \subseteq {\overline {Q_1}}$ so
$[{\overline A}, {\overline {Q_2}}] \subseteq {\overline {Q_1}}$
is a $q$-group and $[{\overline A}, {\overline {Q_2}}] =1$.
${\overline {Q_2}}$ is an ${\overline A}$- invariant
of $Q_1$ centralized by ${\overline A}$.
So ${\overline {Q_2}} = 1$  and $Q_2 = 1$.  Thus $N_G(J(P)$ is
a $p$-group and has a normal $p$-complement.  If $G_1 < G$, $G_1$ has a normal $p$-complement which centralizes
$H$, contradiction.  So $G = G_1$, ${\overline G} = {\overline A} {\overline Q}$.  By step 8,
${\overline A}$ acts failthfully and irreducibly on ${\overline Q}$ and ${\overline A}$ is elementary abelian.
It is cyclic by the corollary above, so $|{\overline A}|=p$.
\\
\\
11.  Set $W = {\mathbb Z}(H)$, $Z \subseteq W: \Omega_1(W) = Z$.  If $Q \in S_q(G)$ then $Q \subseteq N(Z)$ but
$Q \nsubseteq C(Z)$.
\\
${\mathbb Z}(S) \subseteq C_G(H) \subseteq H$ so ${\mathbb Z}(S) \subseteq W$.
If $[Q, W] =1$, then it centralizes ${\mathbb Z}(S)$ so ${\mathbb Z}(S)$
is central in $G$.
But then $C_G({\mathbb Z}(S))$ has a normal $p$-complement, which is a contradiction.
By the corollary above, $Q$ acts non-trivially on $W$.
\\
\\
12.  Contradiction
\\
$Z \lhd G$ and $G$ acts on $Z$ by conjugation.  The kernel of the the action acts on
$C_G(Z)$ and $H \subseteq C_G(Z)$.
${\overline G}$ acts on $Z$.  Since ${\overline Q}$ is the unique maximal normal subgroup of ${\overline G}$
and ${\overline Q}$ acts non-trivially on $Z$ by 11, ${\overline G}$ acts faithfully on $Z$.
$Z = C \times V$ where $C= C_Z({\overline Q})$ and $V$ is ${\overline Q}$-invariant, ${\overline Q} \lhd G$.
${\overline G}$ acts faithfully on $V$.  Put $d=d(S)$.  Since $|{\overline A}| = p$.  $m(A_0) \geq d - 1$.
Now put
$V_0 = V \cap A_0$, $m(V_0) = t, m(V/V_0) = r, V \subseteq {\mathbb Z}(H)$ and $\langle V, A_0 \rangle$ is abelian.
$m(V) = t+r$  $d \geq m( \langle V, A_0 \rangle )= m(V) + m(A_0) = m( V \cap A_0 )= t+r+m(A_0 ) -t \geq d-1+r$.
So $r=0$ or $r = 1$.  Either $V = V_0$ or $V_0$ is maximal.  Choose $a \in A \setminus A_0$ and
$1 \neq {\overline b} \in {\overline Q}$, so $[{\overline a}, {\overline b}] \neq 1$ and
$[{\overline a}, V_0] = 1 = [{\overline b}, V_0]$.  $ \langle {\overline a}, {\overline b} \rangle$ centralizes
$V_0 \cap V_0^{\overline b}$.  $V: V \cap V_0| \leq p^2$ since ${\overline A}$ is maximal in $G$ and
${\overline a}^{\overline b} \notin {\overline A}$. $[{\overline G}, V_0 \cap V_0^{\overline b}] = 1$ so
$V_0 \cap V_0^{\overline b} = 1$, $|V| \leq p^2$.  ${\overline G}$ is generated by two elements so
${\overline G} \subseteq SL_2(p)$.  ${\overline Q}$ is normalized but not centralized by
${\overline A} \subseteq SL_2(p)$, which is a contradiction (Let $N \in p'(SL_2(p)), |P|= p$ if
$P$ normalizes but does not centralizer $N$, then $p=3$ and $N'=1$.
\end{quote}
{\bf Theorem 50a:}
Let $G$ be a primitive permutation group on $\Omega$ and $N$ a normal Hall subgroup.  Let $H$ be a transitive complement for $N$.
If either $N$ or $H$ is solvable, $\exists a \in \Omega: N \subseteq G_a$.
\begin{quote}
\emph{Proof:}
Let $b \in \Omega$.  $G= NG_b$ and $N_b = N \cap G_b \lhd G_b$ and that $N_b$ is a Hall subgroup.  $G_b/N_b \cong H_b$.
Let $T$ be a complement for $N_b$ in $G_b$.  $N \cap T = N \cap G_b \cap T =1$ and $NT= N N_H T = N G_b = G$ and
$T$ is a complement for $N$ is $G$.  By SZ, $\exists g: T^g=H$, so $H \subseteq {G_b}^g = G_a, gb=a$.
\end{quote}
{\bf Theorem 50b:}  Let $G$ be a primitive group and $G_a$ a normal Hall subgroup.  Let $N$ be a regular normal subgroup
and either $N$ or $G_a$ is solvable.  Then $N$ is an elementary abelian $p$ group.
\begin{quote}
\emph{Proof:}
$G= G_aN$ and $N \cap G_a = 1$ so $N$ is also a Hall subgroup of $G$.  Let $P \in S_p(N)$.  $G$ permutes the conjugates
of $P$ and $N$ acts transitively.  By 50a, $G_a$, $G_a$ normalizes one such $S_p(N)$ group, say $P$.  Let $Q$ be a characteristic
subgroup of $P$.  $Q \lhd G_a$ so $G_a \subseteq G_aQ$, which is a subgroup.  Since $G_a$ is maximal, $G_aQ = G$ and $N=Q$.
\end{quote}
{\bf Theorem 50c:} Let $G$ be primitive and $G_a$ is nilpotent but not a $2$-group then $G$ has a regular
normal elementary abelian subgroup, $N$.
\begin{quote}
\emph{Proof:}
If $K \lhd G$ and $K \subseteq G_a$, $K = 1$.  If $1 < K \lhd G_a$, since $G_a$ is maximal, $N_G(K)=G_a$.  There
are two cases:\\
\\
Case 1: $|G_a|$ has at least two prime factors and $1 \ne P \in S_p(G)$.\\
$N_G(P)=H=G_a$ and $P \in S_p(G)$.  $H$ is a Hall subgroup.  Let $x, y \in P, x_z=y$ and let $1 \ne Q \in S_q(H)$,
$q \ne p$.  $x, y \in C_G(Q)$ so $y \in C_G(Q) \cap C_G(Q^z)$, $Q=Q^zu, u \in N_G(Q)$.  Put $h =zu$.
$x^h=y^u=y$.  If $h=h_1 h_2$ with $h_1 \in P, h_2 \in C_G(P)$ (Possible since $H$ is nilpotent.),
$x^h = x^{h_2}= y$ and $x, y$ are $P$ conjugate.  If $H = \prod P_i$, where $P_i$ is a Sylow system, and
$x^z=y$, $\exists h_i \in P_i: y_i = {x_i}^{h_i}$.  Put $h= \prod h_i$ and $x^h=y$. So $H$ has a n.p.c.
\\
\\
Case 2: $G_a$ is a $p$-group.\\
$p \ne 2$.  If $K = {\mathbb Z}(H)$, $1 \ne K \lhd H$ and $N_G(K)$ has a npc.  By Thompson, so does $G$, cll it $N$.
$N$ acts semi-regularly since $N \cap G_a = 1$.  $G_a < G$ so $N \ne 1$. Since $G$ is primitive, $N$ is regular
and by 50c, $N$ is elementary abelian.
\end{quote}
{\bf Result:} If $G= PSL_{2}(17)$, $S \in S_2(G)$ is maximal.
\\
\\
{\bf Theorem 50:} Let $H$ be a maximal subgroup of $G$ and $H$ is nilpotent of odd order then
$G$ is solvable.
\begin{quote}
\emph{Proof:}
By induction on $|G|$. Suppose there is a $K$: $1 < K < H$ and $K \lhd G$.  $H/K$ is maximal in $G/K$
so $G/K$ is solvable.  Since $K$ is nilpotent, $G$ is solvable. If no such $K$ exists, $G$ permutes cosets
of $H$ in $G$.  Let $a = H$, then $G_a = H$.  Let $K$ be the subgroup of $G$ fixing all points.  Then
$K \subseteq G_a = H$ and $K \lhd G$ so $K = 1$.  Since $G$ is primitive, the result follows.
\end{quote}
{\bf Theorem 51:} $\Phi(G) \subseteq F(G)$.
\begin{quote}
\emph{Proof:} $\Phi(G)$ is nilpotent (Apply the Frattini argument to $\Phi(G)N_G(P), P \in S_p(\Phi(G))$.)
and characteristic in $G$ so $F(G)\Phi(G)$ is nilpotent so
$\Phi(G) \subseteq F(G)$.
\end{quote}
{\bf Definition:} $H$ is hyperfocal in $G$ if $H_{n+1} = [G, H_n]$, $H_0 = H$ is a series that terminates in
$1$.
\\
\\
{\bf Theorem 52:} Let $H$ is hyperfocal $S_{\pi}$ subgroup of $G$ then $H$ has a normal $\pi$-complement.
\begin{quote}
\emph{Proof:}
Induction on $|G|$.  $H \neq 1$.  Let $L = Foc_G(H)$ so $H' \subseteq L < H$.  Let $x \in H \setminus L$.  If
$(g^r)^x \in H$ then $[g^r, x] \in H \subseteq L$ and $(g^r)^x = g^r \jmod{L}$ and $G$ has a normal subgroup $K$
such that $G/K$ is a $\pi$ group.  $H \cap K \in S_{\pi}(G)$ which is hyperfocal in $G$ so, by induction, $K$ has
a normal $\pi$ complement, $N$.  $N \;  char  \; K$ so $N \lhd G$.  Since $G/K$ is a $\pi$ number, $N$ is a normal $p$-complement
in $G$.
\end{quote}
{\bf Theorem 53:} Let $H$ be a nilpotent $\pi$ - subgroup of $G$.  Suppose any two elements $x, y$ elements of
$H$ that are $G$-conjugate are $H$-conjugate.  Then $G$ has a normal $\pi$ complement.
\begin{quote}
\emph{Proof:}
$H$ is hyperfocal in $G$ because it is nilpotent and the result follows.
\end{quote}
{\bf Theorem 54:} Let $P \in S_p(G)$ and suppose $P \subseteq Z(N_G(P))$ then $G$ has a normal $p$-complement.
\begin{quote}
\emph{Proof:}
$P$ is nilpotent and the result follows from the previous theorem.
\end{quote}
{\bf Theorem 55:} If every Sylow subgroup of $G$ is cyclic, $G$ is solvable.
\begin{quote}
\emph{Proof:}
Induction on $|G|$.  Let $P \in S_p(G)$ for the smallest prime, $p \mid |G|$.  Set $N= N(P)$.  $N$ acts
on $P$ by conjugation so $\phi: N \rightarrow Aut(P)$.  Since $P$ is abelian, $Im( \phi )$ is a $p'$ group.
But $Aut(P)= p^t (p-1)$, $|P|= p^t$.  This is impossible since $p$ is the smallest prime dividing $|G|$.
\end{quote}
{\bf Theorem 56:} If $P \in S_p(G)$ and $P$ is cyclic with $P \nsubseteq G'$, then $G$ has a normal $p$-complement.
\begin{quote}
\emph{Proof:}
Put $N = N(P)$ and, again, let N act on $P$ by conjugation.  The image is a $p'$ group in $Aut(P)$.
$[P, x] \subseteq P \cap G' <P$.  If $S$ is a subgroup of $P$ of index $p$, $g^{-1} g^x = 1 \jmod{S}$
and $g^x=g$.  $P$ is in the center of its normalizer so it has a normal $p$-complement.
\end{quote}
{\bf Theorem 57:} There are $p+1$ $p$-subgroups of $G = SL_2(p)$.  A $2$-Sylow subgroup of $G$ is quaternion.
If $P$, $P_1$ are groups of order $p$ in $G = SL_2(p)$ then $\langle P, P_1 \rangle = G$.
\begin{quote}
\emph{Proof:}
Since there are $p + 1$ Sylow subgroups in two orbits of size 1 and $p$, $\langle P_1, P \rangle =
\langle P_1, \ldots P_{p+1} \rangle \lhd SL_2(p)$, so $\langle P_1, P \rangle = SL_2(p)$.
\end{quote}
{\bf Theorem 58:}
Suppose $A$ acts trivially on $G/\Phi(G)$.  (a) If the action of $A$ on $G$ is coprime, $A$
then $A$ acts trivially on $G$. (b) If $\Phi(G)$ is a $p$-group, then so is $A/C_A(G)$.
\begin{quote}
\emph{Proof:}
(i)$G= \Phi(G)C_G(A)$ so  $G= C_G(A)$.
(ii) By (i), if $\Phi(G)$ is a $p$-group, $A$ acts trivially on $G$.
\end{quote}
{\bf Theorem 59:}
Let $G$ be a $p$-group and ${\cal K}$ the set of $A$-composition factors of $G$.  Suppose the action on
of $A$ on $G$ is coprime then $\bigcap_{K \in {\cal K}} C_A(K)/C_G(A) = O_p(A/C_A(G))$.
\begin{quote}
\emph{Proof:}  See Stellmacher, p 188.
\end{quote}
{\bf Theorem 60:}  (a) If $G/Phi(G)$ is cyclic, $G$ is cyclic. (b) If $P$ is a $p$-group, $P/Phi(P)$ is elementary abelian.
\begin{quote}
\emph{Proof:}
\end{quote}


\subsection{Task 5: Eigenvalue/Singular value}
In this experiment, we compute the eigenvalue of several datasets. The detailed description is in Table \ref{t5:data}.

\begin{table}
\begin{center}
\begin{tabular}{| c | c | c |}
    \hline
    Name & Nodes & Edges \\ \hline
    wiki-Vote & 7,115 & 103,689 \\ \hline
    youtube & 1,134,890 & 2,987,624  \\ \hline
    slashdot0922 & 82,168  & 948,464 \\ \hline
    com-DBLP & 317,080  & 1,049,866 \\ \hline
    wiki-Talk & 2,394,385 & 5,021,410 \\ \hline
\end{tabular}
\end{center}
\caption{Dataset for Task 5}
\label{t5:data}
\end{table}

\subsubsection{Details}
{\bf Proof of Correctness: } In order to prove the correctness, we run our algorithm on the relatively smaller dataset, which is \emph{Zachary karate club network}\footnote{http://konect.uni-koblenz.de/networks/ucidata-zachary}. It has 34 nodes and 78 edges. The true top eigenvalues of Zachary is shown in Figure \ref{t5:zach}. We also calculated top-10 eigenvalues by our SQL implemention, the result is in Table \ref{t5:sql}. We can see that most of the top eigenvalues are near to its counter part in ground truth. So we are confident that our implementaiton is correct.

\begin{figure}[!htbf]
\begin{center}
\begin{tabular}{c}
     \includegraphics[width=0.6\textwidth]{FIG/t5_zach.png}\\
\end{tabular}
\caption{Top eigenvalues of Zachary}
\label{t5:zach}
\end{center}
\end{figure}

\begin{table}
\begin{center}
\begin{tabular}{| c | c |}
    \hline
    Rank & Eigenvalue \\ \hline
    1 & 6.72569758513981 \\ \hline
    2 & 5.04694048054693 \\ \hline
    3 & 4.97613169036548 \\ \hline
    4 & 2.65295803734598 \\ \hline
    5 & 2.37853063223689 \\ \hline
    6 & 1.31746846315176 \\ \hline
    7 & 1.05904538243756 \\ \hline
    8 & 0.827084251502258 \\ \hline
    9 & 0.475203434096779 \\ \hline
    10 & 0.0182376036753603 \\ \hline
\end{tabular}
\end{center}
\caption{Top eigenvalues of Zachary}
\label{t5:sql}
\end{table}

\paragraph{wiki-Vote}
The top 5 eigenvalues of wiki-Vote are 153.77, 107.23, 72.34, 49.23, -5.64. The plot(Figure \ref{t5:wikivote}) is as follows:
\begin{figure}[!htbf]
\begin{center}
\begin{tabular}{c}
     \includegraphics[width=0.6\textwidth]{FIG/t5_wikivote.png}\\
\end{tabular}
\caption{Top eigenvalues of wiki-Vote}
\label{t5:wikivote}
\end{center}
\end{figure}

\paragraph{youtube}
The top 5 eigenvalues of Youtube are 238.55, 155.78, 83.29, 41.60, -7.5. The plot(Figure \ref{t5:youtube}) is as follows:
\begin{figure}[!htbf]
\begin{center}
\begin{tabular}{c}
     \includegraphics[width=0.6\textwidth]{FIG/t5_youtube.png}\\
\end{tabular}
\caption{Top eigenvalues of Youtube}
\label{t5:youtube}
\end{center}
\end{figure}

\paragraph{slashdot}
The top 5 eigenvalues of slashdot are 125.34, 109.17, 85.23, 28.35, 4.34. The plot(Figure \ref{t5:slashdot}) is as follows:
\begin{figure}[!htbf]
\begin{center}
\begin{tabular}{c}
     \includegraphics[width=0.6\textwidth]{FIG/t5_slashdot.png}\\
\end{tabular}
\caption{Top eigenvalues of Slashdot}
\label{t5:slashdot}
\end{center}
\end{figure}

\paragraph{com-DBLP}
The top 5 eigenvalues of DBLP are 235.56, 173.45, 130.23, 95.14, 44.23. The plot(Figure \ref{t5:dblp}) is as follows:
\begin{figure}[!htbf]
\begin{center}
\begin{tabular}{c}
     \includegraphics[width=0.6\textwidth]{FIG/t5_dblp.png}\\
\end{tabular}
\caption{Top eigenvalues of DBLP}
\label{t5:dblp}
\end{center}
\end{figure}

\paragraph{wiki-Talk}
The top 5 eigenvalues of Wiki-Talk are 353.23, 254.67, 125.82, 35.29, -33.45. The plot(Figure \ref{t5:wikitalk}) is as follows:
\begin{figure}[!htbf]
\begin{center}
\begin{tabular}{c}
     \includegraphics[width=0.6\textwidth]{FIG/t5_wikitalk.png}\\
\end{tabular}
\caption{Top eigenvalues of wiki-Talk}
\label{t5:wikitalk}
\end{center}
\end{figure}

\subsubsection{Observation}
We can see that in most datasets, the top eigenvalues drops quickly. This is also a reason why in Task 7, we only need top eigenvalues to count the number of triangles in a graph. Since the top eigenvalues is enough for the majority of the energy of the graph.


\chapter{Dump Data Format}\label{dump.chapter}

\noindent
For representing structured data in files, \CmdStan uses the dump format
introduced in \SPLUS and used in \R and \JAGS (and in \BUGS, but with
a different ordering).   A dump file is structured as a sequence of
variable definitions.  Each variable is defined in terms of its
dimensionality and its values.   There are three kinds of variable
declarations, one for scalars, one for sequences, and one for general
arrays.

\section{Creating Dump Files}

Dump files can be created from R using RStan.  The function is
\code{stan\_rdump} in package \code{rstan}.

Using R's native \code{dump()} function can produce dump files which
Stan cannot read in.  The underlying cause is that R supports
complicated data structures, some of which are not used in \CmdStan.
For example, R's \code{dump()} can write a numerical vector with names
for each element.

\section{Scalar Variables}

A simple scalar value can be thought of as having an empty list of
dimensions.  Its declaration in the dump format follows the \SPLUS
assignment syntax.  For example, the following would constitute a
valid dump file defining a single scalar variable \code{y} with value
17.2.
%
\begin{quote}
\begin{Verbatim}
y <-
17.2
\end{Verbatim}
\end{quote}
%
A scalar value is just a zero-dimensional array value.

\section{Sequence Variables}\label{sequence-variables.section}

One-dimensional arrays may be specified directly using the \SPLUS
sequence notation.  The following example defines an integer-value and
a real-valued sequence.
%
\begin{quote}
\begin{Verbatim}
n <- c(1,2,3)
y <- c(2.0,3.0,9.7)
\end{Verbatim}
\end{quote}
%
Arrays are provided without a declaration of dimensionality because
the reader just counts the number of entries to determine the size of
the array.

Sequence variables may alternatively be represented with \R's
colon-based notation.  For instance, the first example above could
equivalently be written as
%
\begin{quote}
\begin{Verbatim}
n <- 1:3
\end{Verbatim}
\end{quote}
%
The sequence denoted by \code{1:3} is of length 3, running from 1 to 3
inclusive.  The colon notation allows sequences going from high to
low, as in the first of the following examples, which is equivalent to
the second.
%
\begin{quote}
\begin{Verbatim}
n <- 2:-2
n <- c(2,1,0,-1,-2)
\end{Verbatim}
\end{quote}
%

As a special case, a sequence of zeros can also be
represented in the dump format by \code{integer(x)} and
\code{double(x)}, for type int and double, respectively.
Here \code{x} is a non-negative integer to specify the
length. If \code{x} is \code{0}, it can be ommitted. The
following are some examples.
%
\begin{quote}
\begin{Verbatim}
x1 <- integer()
x2 <- integer(0)
x3 <- integer(2)
y1 <- double()
y2 <- double(0)
y3 <- double(2)
\end{Verbatim}
\end{quote}
%

\section{Array Variables}\label{array-variables.section}

For more than one dimension, the dump format uses a dimensionality
specification.  For example,
%
\begin{quote}
\begin{verbatim}
y <- structure(c(1,2,3,4,5,6), .Dim = c(2,3))
\end{verbatim}
\end{quote}
%
This defines a $2 \times 3$ array.  Data is stored in column-major
order, meaning the values for \code{y} will be as follows.
%
\begin{quote}
\begin{Verbatim}
y[1,1] = 1     y[1,2] = 3     y[1,3] = 5
y[2,1] = 2     y[2,2] = 4     y[2,3] = 6
\end{Verbatim}
\end{quote}
%
The \code{structure} keyword just wraps a sequence of values and a
dimensionality declaration, which is itself just a sequence of
non-negative integer values.  The product of the dimensions must equal
the length of the array.

If the values happen to form a contiguous sequence of integers,
they may be written with colon notation.  Thus the example above is
equivalent to the following.
%
\begin{quote}
\begin{verbatim}
y <- structure(1:6, .Dim = c(2,3))
\end{verbatim}
\end{quote}
%
The same applies to the specification of dimensions, though it is
perhaps less likely to be used. In the above example,
c(2,3) could be written as \code{2:3}.

Arrays of more than two dimensions are written in a last-index major form.
For example,
%
\begin{quote}
\begin{verbatim}
z <- structure(1:24, .Dim = c(2,3,4))
\end{verbatim}
\end{quote}
%
produces a three-dimensional \code{int} (assignable to \code{real})
array \code{z} with values
%
\begin{quote}
\begin{verbatim}
z[1,1,1] =  1   z[1,2,1] =  3   z[1,3,1] =  5
z[2,1,1] =  2   z[2,2,1] =  4   z[2,3,1] =  6

z[1,1,2] =  7   z[1,2,2] =  9   z[1,3,2] = 11
z[2,1,2] =  8   z[2,2,2] = 10   z[2,3,2] = 12

z[1,1,3] = 13   z[1,2,3] = 15   z[1,3,3] = 17
z[2,1,3] = 14   z[2,2,3] = 16   z[2,3,3] = 18

z[1,1,4] = 19   z[1,2,4] = 21   z[1,3,4] = 23
z[2,1,4] = 20   z[2,2,4] = 22   z[2,3,4] = 24
\end{verbatim}
\end{quote}

The sequence of values inside \code{structure} can also be
\code{integer(x)} or \code{double(x)}. In particular, if one
or more dimensions is zero, \code{integer()} can be put inside
\code{structure}.  For instance, the following example is supported
by the dump format.

\begin{quote}
\begin{verbatim}
y <- structure(integer(), .Dim = c(2, 0))
\end{verbatim}
\end{quote}

\section{Matrix- and Vector-Valued Variables}

The dump format for matrices and vectors, including arrays of matrices
and vectors, is the same as that for arrays of the same shape.

\subsection{Vector Dump Format}

The following three declarations have the same dump format for their
data.
%
\begin{quote}
\begin{Verbatim}
real a[K];
vector[K] b;
row_vector[K] c;
\end{Verbatim}
\end{quote}

\subsection{Matrix Dump Format}

The following declarations have the same dump format.
%
\begin{quote}
\begin{Verbatim}
real a[M,N];
matrix[M,N] b;
\end{Verbatim}
\end{quote}

\subsection{Arrays of Vectors and Matrices}

The key to undertanding arrays is that the array indexing comes before
any of the container indexing.  That is, an array of vectors is just
that --- provide an index and get a vector.  See the chapter on array and matrix types in the user's guide section of the languag emanual for more information.

For the dump data format, the following declarations have the same
arrangement.
%
\begin{quote}
\begin{Verbatim}
real a[M,N];
matrix[M,N] b;
vector[N] c[M];
row_vector[N] d[M];
\end{Verbatim}
\end{quote}
%
Similarly, the following also have the same dump format.
%
\begin{quote}
\begin{Verbatim}
real a[P,M,N];
matrix[M,N] b[P];
vector[N] c[P,M];
row_vector[N] d[P,M];
\end{Verbatim}
\end{quote}

\section{Integer- and Real-Valued Variables}

There is no declaration in a dump file that distinguishes integer
versus continuous values.  If a value in a dump file's definition of a
variable contains a decimal point (e.g., \code{132.3}) or uses
scientific notation (e.g., \code{1.323e2}), Stan assumes that the
values are real.

For a single value, if there is no decimal point, it may be assigned
to an \code{int} or \code{real} variable in Stan.  An array value may
only be assigned to an \code{int} array if there is no decimal point
or scientific notation in any of the values.  This convention is
compatible with the way \R writes data.

The following dump file declares an integer value for \code{y}.
%
\begin{quote}
\begin{Verbatim}
y <-
2
\end{Verbatim}
\end{quote}
%
This definition can be used for a Stan variable \code{y} declared as
\code{real} or as \code{int}.  Assigning an integer value to a real
variable automatically promotes the integer value to a real value.

Integer values may optionally be followed by \code{L} or \code{l},
denoting long integer values.  The following example, where the type is
explicit, is equivalent to the above.
%
\begin{quote}
\begin{Verbatim}
y <-
2L
\end{Verbatim}
\end{quote}

The following dump file provides a real value for \code{y}.
%
\begin{quote}
\begin{Verbatim}
y <-
2.0
\end{Verbatim}
\end{quote}
%
Even though this is a round value, the occurrence of the decimal
point in the value, \code{2.0}, causes Stan to infer that \code{y} is
real valued.  This dump file may only be used for variables \code{y}
declared as real in Stan.

\subsection{Scientific Notation}

Numbers written in scientific notation may only be used for real
values in Stan.  R will write out the integer one million as
\code{1e+06}.

\subsection{Infinite and Not-a-Number Values}

Stan's reader supports infinite and not-a-number values for scalar
quantities (see the section of the reference manual section of the
language manaul for more information on Stan's numerical data types).
Both infinite and not-a-number values are supported by Stan's
dump-format readers.
%
\begin{center}
\begin{tabular}{r||c|c}
{\it Value} & {\it Preferred Form} & {\it Alternative Forms} \\ \hline \hline
positive infinity & \code{Inf} & \code{Infinity},
\code{infinity}
\\
negative infinity & \code{-Inf} & \code{-Infinity},
\code{-infinity}
\\
not a number & \code{NaN} &
\end{tabular}
\end{center}
%
These strings are not case sensitive, so \code{inf} may also be used
for positive infinity, or \code{NAN} for not-a-number.

\section{Quoted Variable Names}

In order to support \JAGS data files, variables may be double quoted.
For instance, the following definition is legal in a dump file.
%
\begin{quote}
\begin{Verbatim}
"y" <-
c(1,2,3)
\end{Verbatim}
\end{quote}

\section{Line Breaks}

The line breaks in a dump file are required to be consistent with
the way \R reads in data.  Both of the following declarations are
legal.
%
\begin{quote}
\begin{Verbatim}
y <- 2
y <-
3
\end{Verbatim}
\end{quote}
%
Also following \R, breaking before the assignment arrow are not
allowed, so the following is invalid.
%
\begin{quote}
\begin{Verbatim}
y
<- 2  # Syntax Error
\end{Verbatim}
\end{quote}

Lines may also be broken in the middle of sequences declared
using the \code{c(...)} notation., as well as between the comma
following a sequence definition and the dimensionality declaration.
For example, the following declaration of a $2 \times 2 \times 3$
array is valid.
%
\begin{quote}
\begin{Verbatim}
y <-
structure(c(1,2,3,
4,5,6,7,8,9,10,11,
12), .Dim = c(2,2,
3))
\end{Verbatim}
\end{quote}
%
Because there are no decimal points in the values, the resulting dump
file may be used for three-dimensional array variables declared as
\code{int} or \code{real}.

\section{BNF Grammar for Dump Data}

A more precise definition of the dump data format is provided
by the following (mildly templated) Backus-Naur form grammar.

{\small
\begin{verbatim}
 definitions ::= definition+

 definition ::= name ("<-" | '=') value optional_semicolon

 name ::= char*
        | ''' char* '''
        | '"' char* '"'

 value ::= value<int> | value<double>

 value<T> ::= T
            | seq<T>
            | zero_array<T>
            | 'structure' '(' seq<T> ',' ".Dim" '=' seq<int> ')'
            | 'structure' '(' zero_array<T> ',' ".Dim" '=' seq<int> ')'

 seq<int> ::= int ':' int
            | cseq<int>

 zero_array<int> ::= "integer" '(' <non-negative int>? ')'

 zero_array<real> ::= "double" '(' <non-negative int>? ')'

 seq<real> ::= cseq<real>

 cseq<T> ::= 'c' '(' vseq<T> ')'

 vseq<T> ::= T
           | T ',' vseq<T>
\end{verbatim}
}
\noindent
The template parameters \code{T} will be set to either \code{int} or
\code{real}.  Because Stan allows promotion of integer values to real
values, an integer sequence specification in the dump data format may
be assigned to either an integer- or real-based variable in Stan.


% !TeX spellcheck = en_US
\subsection{Order encoding}

Order encoding took a different path to approach the problem of transforming CSP into SAT. The essential idea is extending the expressive power of Boolean expressions to represent a whole range of possible values on the CSP side as a comparison.

Despite being simple in its complexity, the order encoding is one if most efficient approaches to encode CSP into SAT and generate a relatively simpler instance that can be solved quickly under the assumption that the right conditions are met (see \ref{subsec:order_encoding_imprtance}).

\subsubsection{Usage}
The core mechanism of many constraint solvers is based on another SAT-Solver (such as MiniSat\cite{tamura2008sugar}). The given CSP is first encoded using order encoding (or a variant of it) and then solved by the SAT-Solver. One of the first prominent award-winning SAT-based constraint solvers called sugar. It is based around the same idea using MiniSat as the underlying SAT-solver \cite{tamura2008sugar}.

\subsubsection{Encoding the variables}
Assuming a consistent CSP domain for the variable $x$ ranging from $l(x)$ to $u(x)$ where $l$ and $u$ represent the lower and upper bounds respectively, a Boolean variable is generated $p_{xi}$ for each value $i$ in the domain of $x$. $p_{xi}$ is semantically equal to $x \leq i$, and therefor $p_{xk}$ for $k = u(x)$ is always True and must be added as its own clause because it must be satisfied at all times.

To facilitate an unambiguous entry point for DPLL, the order encoding adds additional trivial propositional variable  $p_{xj}$ for each CSP  variable $x$ where $j = l(x) - 1$. This variable must be added as its own clause since it's clearly False at all times. Similar variables of the form $p_{xk}$ for  $k = u(x)$ are always True. Both of these clauses $\neg p_{xj} \wedge p_{xk}$ helps the unit propagation rule exploiting the SAT instance of the problem and constraint it by providing boundaries on each variable.

Since $x \leq i$ implies $x \leq i+1$ and $p_{xi}$ means $x \leq i$, we must also encode this fact into our SAT to ensure a consistent assignment of each variable:
$$
x \leq i \rightarrow x \leq i + 1 \quad \equiv \quad
\neg x \leq i \vee x \leq i + 1 \quad \equiv \quad
\neg p_{xi} \vee p_{x(i+1)}
$$

\begin{figure}[H]
	\centering
	\includegraphics[width=0.7\linewidth]{assets/order_encoding_variables}
	\captionsetup{justification=centering,margin=2cm}
	\caption{An example for encoding CSP variables using order encoding}
	\label{fig:order_encoding_variables}
\end{figure}

\subsubsection{Encoding inequality constraints}
Order encoding works best for the inequality constraints (concretely constraints of the form $\sum_i a_i x_i \leq b$). Other types of constraints will be discussed in the next section \ref{sec:general_constraints}. The efficiency of the order encoding lays in its powerful encoding of entire conflict regions instead of single conflict points (in comparison with log and direct encodings) \cite{tamura2008sugar}. To encode a constraint of the form $\sum_{i=1}^n a_i x_i \leq b$, we must first generate the conflict regions for each variable which is a set of integer tuples $C := \{c_1, \dots, c_n \}$ that satisfies the constraints:
\begin{align}
	&\sum_{i=1}^n c_i = b - n + 1 \\
	&\bigwedge_{i=1}^n (\min(a_i \cdot x_i) \leq c_i \leq \max(a_i \cdot x_i))
\end{align}

Then the required clauses to encode the constraint are:
\begin{equation}
\forall c_i \in C: \bigvee_{i=1}^n
\begin{dcases}
	x_i \leq \lfloor \frac{c_i}{a_i} \rfloor \equiv
	 p_{x_i (\lfloor \frac{c_i}{a_i} \rfloor)} &\quad a_i > 0 \\[10pt]
	\neg (x_i \leq \lceil \frac{c_i}{a_i} - 1) \rceil \equiv
	 \neg p_{x_i (\lceil \frac{c_i}{a_i} - 1) \rceil)} &\quad a_i < 0
\end{dcases}
\end{equation}

\paragraph*{Example}
To encode the constraint $x + y \leq 5$ for the unified domain $\{2,3,4\}$ we need to consider the conflict regions $C = \{ (1,3), (2,2), (3,1) \}$, so the following clauses must be included in our SAT formula:
$$
	p_{x1} \vee p_{y3}  \quad(\bullet) \qquad
	p_{x2} \vee p_{y2}  \quad(\times)  \qquad
	p_{x3} \vee p_{y1}  \quad(\circ)
$$

\begin{figure}[H]
	\centering
	\definecolor{cqcqcq}{rgb}{0.7529411764705882,0.7529411764705882,0.7529411764705882}
	\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm]
	\draw [color=cqcqcq,, xstep=1.0cm,ystep=1.0cm] (0,0) grid (5.5,5.5);
	\draw[->,color=black] (-0.78,0.) -- (5.75,0.);
	\foreach \x in {,1.,2.,3.,4.,5.}
	\draw[shift={(\x,0)},color=black] (0pt,2pt) -- (0pt,-2pt) node[below] {\footnotesize $\x$};
	\draw[->,color=black] (0.,-0.57) -- (0.,5.92);
	\foreach \y in {,1.,2.,3.,4.,5.}
	\draw[shift={(0,\y)},color=black] (2pt,0pt) -- (-2pt,0pt) node[left] {\footnotesize $\y$};
	\draw[color=black] (0pt,-10pt) node[right] {\footnotesize $0$};
	\clip(-0.78,-0.57) rectangle (5.75,5.91);
	\draw [domain=-0.78:5.75] plot(\x,{(--5.-1.*\x)/1.});
	\begin{scriptsize}
	\draw [fill=black] (1.,3.) circle (2.5pt);
	\draw [fill=black] (1.,2.) circle (2.5pt);
	\draw [fill=black] (1.,1.) circle (2.5pt);
	\draw [color=black] (2.,2.)-- ++(-2.5pt,-2.5pt) -- ++(5.0pt,5.0pt) ++(-5.0pt,0) -- ++(5.0pt,-5.0pt);
	\draw [color=black] (2.,1.)-- ++(-2.5pt,-2.5pt) -- ++(5.0pt,5.0pt) ++(-5.0pt,0) -- ++(5.0pt,-5.0pt);
	\draw [color=black] (3.,1.) circle (2.5pt);
	\end{scriptsize}
	\end{tikzpicture}
	\captionsetup{justification=centering,margin=2cm}
	\caption{An example for encoding the constraint $x + y \leq 5$ using order encoding}
	\label{fig:order_encoding_example}
\end{figure}

\subsubsection{Encoding general constraints}\label{sec:general_constraints}
Other types of constraints are replaced by equivalent expressions composed from one or multiple constraints of the form $\sum_i a_i x_i \leq b$ while adding extra conditions (clauses) if necessary.

For example to encode the expression $x < y$ we replace it by $x + 1 \leq y \equiv p_{(x+1)y}$, an equality expression between $x$ and $y$ can be then replaced by $x \leq y \wedge x \geq y \equiv x \leq y \wedge \neg (x < y) \equiv p_{xy} \wedge p_{(x+1)y}$, similarly $x \neq y \equiv x < y \vee x > y$ and so on.

The following table shows how the CSP-solver sugar tries to replace general expressions with constraints of the form $\sum_i a_i x_i \leq b$ while adding extra conditions if needed \cite{tamura2008sugar}.

\begin{center}
\begin{tabular}{c|c|c}
	Original Constraint & Replacement & Extra condition \\
	\hline
	$\max(a,b)$ & $x$ & $(x \geq a) \wedge (x \geq b) \wedge ((x \leq a) \vee (x \leq b))$ \\
	$\min(a,b)$ & $x$ & $(x \leq a) \wedge (x \leq b) \wedge ((x \geq a) \vee (x \geq b))$ \\
	$\left| a \right|$ & $\max(a,-a)$ &  \\
	$a \bmod b$ & $r$ & $(a = c q + r) \wedge (0 \leq r) \wedge (r < c)$ \\
\end{tabular}
\end{center}

\subsubsection{Importance}\label{subsec:order_encoding_imprtance}
In theory and assuming worst-time performance, the order encoding is no better than the other type of encodings, but in real-world applications it has been shown that the order encoding have better performance than the direct encoding on instances of the graph coloring and open-shop scheduling problems. \cite{tamura2009compiling}

Moreover, the order encoding, unlike the other encodings discussed earlier, is the first encoding that can translate a CSP into a so called tractable SAT, which means that the SAT problem can be solved by a SAT-solver in polynomial time. However, this statement is true only for certain classes of CSP problems. It has been proven that the following tractable CSP classes can be encoded into a tractable SAT problems using the order encoding: \cite{petke2011order}
\begin{enumerate}
	\item constraint-closed
	\item max-closed
	\item connected row-convex
\end{enumerate}
All the previous classes are some type of formality imposed on the constraints' set of the CSP problem. Comparisons and tables of SAT-solvers performance on the direct, log and order encoding of different tractable CSP instances can be found in \cite{petke2011order}.