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\begin{document}

\lecture{\ 2}{V. Arvind}{V. Arvind}{Ramprasad Saptharishi}

\section{Motivation}

Last lecture we had a brush up of group theory to set up the arsenal
required to study Graph Isomorphism. This lecture we shall see how
group theory motivates graph isomorphism, and some more theorems on
group theory that we would require for later lectures.

\section{Graph Isomorphism and Automorphism Groups}

Recall that two graphs $G_1$ and $G_2$ are isomorphic if there is a
re-numbering of vertices of one graph to get the other, or in other
words, there is an automorphism of one graph that sends it to the
other. 

And clearly, $Aut(G) \leq S_n$, the symmetric group on $n$ objects,
which represent the permuation group on the vertices. And since it is
a subgroup of the permutation group, $|Aut(G)| \leq n!$

Of course, providing the entire automorphism group as output would
take exponential time but what about a small generating set? Which
then leads us to, does there exist a small generating set?

\begin{lemma}
For any group $H$ of size $n$, there exists a generating set of size
$\log n$. 
\end{lemma}
\begin{proof}
Let $H_0 = \{e\}$. If $H_0 = H$ we are done. Otherwise, let $x = H
\setminus H_0$. Let $H_1 = \langle H_0,x\rangle$ and in general, $x
\in H \setminus H_i$ and $H = \langle H_i, x\rangle$ and since $x$
forms two distinct cosets of $H_i$, $|H_{i+1}| = 2|H_i|$. And hence,
in $\log n$ steps one would hit $H$. 
\end{proof}

Now we can ask the question, can we output a small generating set of
the automorphism group of a graph $G$? We shall refer to this problem
as $Graph-Aut$. We shall now show that $Graph-Iso$ and $Graph-Aut$ are
polynomial time equivalent. 

\begin{theorem}
With $Graph-Iso$ as an oracle, there is a polynomial time algorithm
for $Graph-Aut$ and vice-versa.
\end{theorem}
\begin{proof}
  First we shall show that we can solve $Graph-Iso$ with $Graph-Aut$
  as an oracle. We are given two graphs $G_1$ and $G_2$ and we need to
  create a graph $G$ using the two such that the generating set of the
  automorphism should tell us if they are isomorphic or not.
  
  Let $G = G_1 \cup G_2$. Suppose additionally we knew that $G_1$ and
  $G_2$ are connected, then a single oracle query would be
  sufficient. If any of the generators of $Aut(G)$ interchanged a
  vertex in $G_1$ with one in $G_2$, then connnectivity should force
  $G_1 \cong G_2$. 
  \medskip

  But what if they are not connected? We then have this vey neat
  trick, $G_1 \cong G_2 \Longleftrightarrow \overline{G}_2 \cong
  \overline{G}_2$, and either $G_1$ or $\overline{G}_1$ has to be
  connected and hence one can check for connectivity and then ask the
  appropriate query.
  \bigskip

  The other direction is a bit more involved. The idea is to see that
  any group is a union of cosets. Hence, suppose
  $$
  H = a_1K\cup a_2K \cup \cdots a_nK
  $$
  then $\{a_1, a_2, \ldots, a_n\}$ along with a generating set for $K$
  form a generating set for $H$. Hence once we have a subgroup $K$
  with small index, we can then recurse on $K$. 

  Hence we are looking for a tower of subgroups
  $$
  Aut(G) = H \geq H_1 \geq H_2 \geq \cdots \geq H_m = \{e\}
  $$
  such that $[H_i:H_{i+1}]$ is polynomially bounded.
  \medskip

  For our graph $G$, let $Aut(G) = H \leq S_n$. We shall use
  Weilandt's notation where $i^\pi$ denotes the image of $i$ under
  $\pi$. In this notation, composition becomes simpler: $(i^\pi)^\tau
  = i^{\pi\cdot \tau}$. 

  Define $H_i = \{ \pi \in H: 1^\pi = 1, 2^\pi = 1, \cdots i^\pi =
  1\}$. And this gives the tower
  $$
  H_0 = H \geq H_1 \geq H_2 \geq \cdots \geq H_{n-1} = \{e\}
  $$
  with the additional property that $[H_i:H_{i+1}]\leq n-i$ since
  there are atmost $n-i$ places to go to when the first $i$ are fixed
  by $H_i$. 

  Now look at the tablaeu 

  % picture
  {\em Picture supposed to come here, needs to be completed}

  \medskip
  
\end{proof}

\section{The Set Stabilizer Problem}

{\bf The Problem Statement:} $H\leq S_n$, given by a small generating
set. Also given is a subset $\Delta \subseteq [n]$. Find the 
$$
stab_{\Delta}(H) = \{ \pi \in H: \Delta^{\pi} =
\Delta\}
$$
Though this problem has nothing to do with graphs directly, graph
isomorphism reduces to this. 

\begin{theorem}
$$Graph-Iso \leq_P Set-Stab$$
\end{theorem}
\begin{proof}

By our earlier theorem, it is enough to show that $Graph-Aut$ reduces
to $Set-Stab$. \medskip

One simply needs to note that an automorphism can be thought of as
acting on the edges as well. Given a graph $G = (V,E)$, a permutation
of the vertices induces a permutation of the edges. Hence, 
$$
\phi: Sym(V) \longrightarrow Sym\binom{V}{2}
$$
is injective. 

Thus, all we need to do is find the set of elements in $Sym(V)$ that
stabilizes $E$. Our set $H$ is $\phi(Sym(V)) \subseteq
Sym\binom{V}{2}$ and $\Delta = E \subseteq \left[\binom{n}{2}\right]$
and the automorphism group is precicely $stab_{\Delta}(H)$. 
\end{proof}

\section{More Group Theory: Sylow Theorems}

We will need some more tools for the lectures that follow, the Sylow
Theorems in particular. Before that, we need the Orbit-Stabilizer
theorem. 

\begin{definition}
Let $G$ act on a set $S$. Let $s\in S$
\begin{itemize}
\item The orbit of $s$ ($s^G$), is the set of all possible images of
  $s$ under the action of $G$. 
$$  s^G  =  \{ t\in S: \exists g\in G, gs = t\}$$

\item The stabilizer of $s$($G_s$) is the set of all elements of $G$
  that fix $s$
$$G_s  =  \{ g\in G: gs = s\}$$
\end{itemize}

\end{definition}

\begin{theorem}[Orbit-Stabilizer Theorem]
For any finite group $G$ that acts on a set $S$. For every $s\in S$, 
$$
|G| = |s^G|\cdot|G_s|
$$
\end{theorem}
\begin{proof}
This is just Lagrange's theorem, all we need to see is that the
stabilizer $G_s$ is a subgroup of $G$ and that $[G:G_s] =
|s^G|$, should be a trivial exercise to the reader. 
\end{proof}


\begin{theorem}[Sylow Theorems]
  Let $G$ be a group, $|G| = p^mr$, where $p$ is a prime and
  $gcd(r,p)=1$. Then
  \begin{enumerate}
  \item there exists a subgroup $P$ such that $|P| = p^m$ ($p$-sylow subgroup)
  \item for any $p$-subgroup\footnote{subgroup of order $p^a$ for some
      $a$}$H$ of $G$, one of its conjucates is contained in $P$
  \item the number of $p$-sylow subgroups of $G$ is $1\pmod{p}$
  \end{enumerate}
\end{theorem}


\begin{proof}
We shall prove the subdivisions one after another. \\

\noindent
{\em Subdivision $1$:}

Let $\Omega$ be the set of subsets of $G$ of size $p^m$. Note that
$|\Omega| = \binom{p^mr}{p^m}$. Lucas' theorem tells us that
$\binom{p^mr}{p^m}$ is not divisible by $p$ by our choice of $r$. Let
$G$ act on $\Omega$ by left multiplication. 

The action decomposes $\Omega$ into orbits, and since $p\nmid
|\Omega|$, there exists $A\in \Omega$ such that $p\nmid |A^G|$. And
since $p^mr|G| = |A^G||G_A|$, and by the choice of $A$, $p^m\mid
|G^A|$. 

And since the elements $ga \in A$ for $a\in A$ are distinct under the
action of $G_A$, it follows that $|A| \geq |G_A|$ and hence forcing
$|G_A| = p^m$. Hence $G_A$ is our desired $p$-sylow subgroup. 
\bigskip

\noindent
{\em Subdivision $2$:}

Let $\Omega$ be the set of left cosets of $P$, our $p$-sylow
subgroup. And let $H$ be a $p$-subgroup of $G$, which induces an
action on $\Omega$ by left multiplication. 

Since $|H| = p^a$, every non-trivial orbit of $\Omega$ is has
cardinality a multiple of $p$. Hence, the number of points of $\Omega$
that are fixed by $H$ is modulo $p$ is the same as $|\Omega|$. Hence
in particular, since $p\nmid |\Omega|$, the set of points fixed by $H$
is non-zero. Hence, there exists a $gP$ that is fixed by $H$. 

Hence $hgP \subseteq gP$ or $g^{-1}hgP \subseteq P \implies g^{-1}hg
\in P$ for all $h\in H$. Thus $g^{-1}Hg \subseteq P$

Note that this also tells us that all $p$-sylow subgroups are
conjucates of each other. 

\bigskip

\noindent
{\em Subdivision $3$:}

Let $P$ be a $p$-sylow subgroup and let $\Omega$ be the set of
$p$-sylow subgroups of $G$ on which $P$ acts by conjucation. For any
$Q\in \Omega$, the stabilizer of $Q$ under conjucation is called the
normalizer of $Q$, denoted by $N_G(Q)$. 

Suppose $Q\in \Omega$ is fixed by $P$ on conjucation, then $P\leq
N_G(Q)$. But subdivision $2$ tells us that $P$ and $Q$ are conjucate
to each other in $N_G(Q)$, which then forces $P = Q$. Hence the only
fixed point is $P$ itself and hence $|\Omega| = 1\pmod{p}$.
\end{proof}

{\em We also did the proof of Lucas' Theorem, has to be \TeX-ed out}
\end{document}

