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\begin{document}
\lecture{\ 3: Divide-and-Conquer on Groups}{V. Arvind}{V. Arvind}{Ramprasad Saptharishi}
\section{Overview}
Last lecture we studied automorphism groups motivated through graph
isomorphism. This lecture we shall examine other techniques to study
group theoretic properties; we shall implement a 'divide-and-conquer'
approach to study groups.
\section{Orbit Computation}
Computing the orbit of an element is one of the basic questions in
group theoretic algorithms.
We are given a group $G$ that acts on a finite set $\Omega$, and hence
$G$ can be thought of as a subgroup of $Sym(\Omega)$. And of course
since $G$ could be very large, a small generating set $A$ of $G$ is
given as input. Given an element $\alpha\in \Omega$, we would like to
compute the $G$-orbit of $\alpha$, denoted by $\alpha^G$. Recall that
$$
\alpha^G = \setdef{\beta \in \Omega}{\exists g\in G, \beta = \alpha^g}
$$
The naive way is to try and 'reach' every element in the orbit using
elements of $A$ acting on $\alpha$, which is as follows:
\medskip
\begin{algorithmic}[1]
\STATE $\Delta = \{\alpha\}$
\WHILE{$Delta$ grows}
\FOR{each $a\in A$ and each $\delta\in D\in\Delta$}
\STATE $\Delta = \Delta \cup \{\delta^g\}$
\ENDFOR
\ENDWHILE
\end{algorithmic}
\medskip
The running time is atleast quadratic in the $|\Omega|$. A
reachability approach is much better. \medskip
\noindent
{\em Algorithm:} Define a graph $X = (V,E)$ where $V = \Omega$ and $(\alpha, \beta)\in
E$ if $\alpha^a = \beta$ for some $a$ in $A\cup A^{-1}$. The connected
components of this graph would correspond to the orbits.
\medskip
This is a linear time algorithm to compute orbits of $\Omega$. Note
that this algorith infact gives more, given any two elements of the
same orbit, one could also obtain a group element that takes one to
another: look at the path from one to another, multiply the edge
labels.
\medskip
This gives us a step forward towards a divide-and-conquer approach.
Once we have the orbit decomposition of $\Omega$, we can study the
action of the group on each orbit separately. We would then have the
additional property of the action of $G$ being transitive\footnote{any
element of the set can be pushed to any other by some element of the
group} on the set.
\section{Decomposition of Transitive Groups}
In order to break down transitive groups we shall study ``blocks''. In
this section, we shall assume that the action of $G$ is transitive.
\begin{definition}
$\Delta \subseteq \Omega$ is called a block if for all $g\in G$,
either $\Delta^g = \Delta$ or $\Delta^g \cap \Delta = \phi$
\end{definition}
Of course for any action, $\Omega$ and singletons of $\Omega$ are
blocks are trivial to be interesting.
\begin{definition}
$G$ is said to be primitive if it has only trivial blocks
\end{definition}
Let us look at some examples.
\begin{enumerate}
\item When $G$ is infact the entire $Sym(\Omega)$, it's easily seen
that $G$ is primitive.
\item Even when $G$ is the set of even permuations over $\Omega$, also
denoted by $A_{|\Omega|}$, it has enough permuations to still remain
primitive. This also can be easily checked.
\item Suppose $G$ acts on itself, say by left multiplication, $G\leq
Sym(G)$. Every subgroup of $G$ is a block since the cosets are
either identical or disjoint. Infact, any coset would also be a
block.
Thus, we have a tiling of $\Omega$ using such blocks, we shall refer
to these as a {\em block system}
\begin{definition}
A block system is a partition of $\Omega$ such that each part in
the partition is a block with respect to the action.
\end{definition}
Note that if $\Delta$ is a block, so is $\Delta^g$ for every $g\in
G$. And the tiling gives a theorem very similar to Lagrange's
theorem on subgroups of a group.\footnote{note that you require the
action to be transitive, otherwise such $\Delta^g$ blocks needn't
cover all of $\Omega$}
\begin{theorem}
Let $G$ act transitively on a set $\Omega$ and let
$\Delta\subseteq \Omega$ be any block. Then $|\Delta|$ divides $|\Omega|$
\end{theorem}
And this immediately leads to the following corollary.
\begin{corollary}
For any group $G$ that acts on $\Omega$ transitively, if
$|\Omega|$ is a prime, then $G$ is primitive.
\end{corollary}
\item Let $X$ be the graph, a collection of some $k$ triangles and
look at its automorphism group acting on it. First note that this
action is transitive, and further, each triangle would be a block.
Thus the automorphism group is imprimitive. \medskip
This can of course be extended to any collection of $k$ identical
graphs such that the automorphism group of the piece is transitive.
\item Look at the leaves of a complete binary tree of depth $k$, and
let the group be the automorphism group acting on them.
What are the blocks?
Take any internal node in the tree, and look at the set of all
descendant leaves of it, and this set of leaves form a block.
Infact, all blocks are precisely such sets of leaves.
\end{enumerate}
The last example infact gives a great motivation to a
divide-and-conquer approach.
If $G$ is transitive and imprimitive, let $\Delta$ be the smallest
block. Now group elements of $\Omega$ corresponding to the
block-system generated by $\Delta$. Now notice that $G$ infact acts on
this block system since $G$ moves the blocks in the system around. And
let the block system be the new set $\Omega_1$ and the group being the
projected version of $G$ and we can now ask the question ``Is $G'$
primitive/transitive ?'' with respect to the smaller set $\Omega$.
Checking if the action is transtive can be done by orbit computation,
but we need to check if a group is primitive.
\section{Blocks and Subgroups}
{\bf Observation: } If $\Delta_1$ and $\Delta_2$ are $G$-blocks, then
so is $\Delta_1\cap \Delta_2$.
With this observation, we can now talk about the smallest block
containing a bunch of elements of $\Omega$.
\begin{lemma}
$G\leq Sym(\Omega)$ acting transitively on $\Omega$, is primitive if
and only if $G_\alpha$ is a maximal subgroup of $G$.
\end{lemma}
\begin{proof}
Note that $\alpha$ needn't be specified since $G_\alpha$ and
$G_\beta$ are conjucates of each other when $G$ is transitive. It
is easy to check that of $g\in G$ such that $g\alpha = \beta$, then
$gG_\alpha g^{-1} = G_\beta$. \smallskip
Suppose $\{\alpha\} < \Delta < \Omega$, a non-trivial block. Let $H
= \setdef{g\in G}{\Delta^g = \Delta}$. We will now show that
$G_\alpha < H < G$, thus proving one direction of the lemma.
Clearly, since $G$ acts transitively and $\Delta < \Omega$, $H$ has
to be a proper subgroup of $G$. Also, if $g\in G_\alpha$, then
$\alpha\in \Delta \cap \Delta^g \neq \phi$. Since $\Delta$ is a block, this
forces $\Delta = \Delta^g$ and thus $g\in H$. Since $\{\alpha\} < \Delta$
there exists a $\beta\in \Delta$ different from $\alpha$. Let $g$ be
the element of the group that takes $\alpha$ to $\beta$. Then since
$\beta \in \Delta\cap \Delta^g$, $g \in H$ but $g\notin
G_\alpha$. Thus $G_\alpha < H$. \\
As for the other direction, let $G_\alpha < H < G$. We shall show
that $\alpha^H = \Delta$ is our non-trivial block. Since $G_\alpha <
H$, $\Delta \neq \{\alpha\}$. Showing $\Delta < \Omega$ is a bit
more involved. Since $G_\alpha$ is a subgroup of $G$, $G_\alpha$ and
its cosets partition $G$:
$$
G = \bigcup_{\beta\in \Omega\ ,\ g_\beta:\alpha\mapsto
\beta}G_\alpha g_\beta
$$
And note that if any $g_\beta \in H$, the the entire coset of
$G_\alpha g_\beta$ is contained in $H$. Hence since $\Delta =
\Omega$ would imply $H = G$, our assumption $H < G$ forces $\Delta <
\Omega$.
All that's left to show is that $\Delta$ is a block. Suppose
$\Delta^g \cap \Delta \neq \phi$, then for some $h,h' \in H$,
$\alpha^{hg} = \alpha^{h'}$ which then forces $h'gh^{-1} \in
G_\alpha < H$. Hence $g\in H$ and therefore $\Delta^g = \Delta$.
\end{proof}
What the above lemma established is a one-to-one correspondence
between subgroups of $G$ and blocks of $\Omega$.
One could also think of $G$ as acting on $\setdef{G_\alpha g}{\alpha
\in \Omega}$, by identifying each point $\alpha \in \Omega$ by the
subgroup $G_\alpha$ and its cosets. \medskip
\begin{lemma}
Let $N\lhd G$, a normal transitive subgroup of $G$. Then the orbits
of $N$ form a block system.
\end{lemma}
\begin{proof}
We want to show that $\alpha^N$ is a block. Suppose $\alpha^{Ng}
\cap \alpha^{N} \neq \phi$, then for some $n_1, n_2 \in N$,
$\alpha^{n_1g} = \alpha^{n_2}$ and hence $n_1gn_2^{-1} \in
G_\alpha$. By normality of $N$, the above terms can be written as
$n_3g$ for some $n_3\in N$. From this, $n_1gn_2^{-1}\in Ng$ and
hence $n_2\in Ng$ which collapses $Ng$ and $N$, thus forcing
$\alpha^{Ng} = \alpha^N$.
\end{proof}
\begin{corollary}
If $G$ is primitive, all its normal subgroups are transitive.
\end{corollary}
\section{Finding Blocks}
{\bf Problem:} Given $\langle A \rangle = G\leq Sym(\Omega)$ a
transitive group. Find a non-trivial block system or report {\sc
primitive}. \medskip
Observe that if $G$ is not primitive, for every $\alpha\in \Omega$,
there exists a $\beta\neq \alpha$ such that $G$ has a non-trivial
block containing $\{\alpha, \beta\}$. And hence it is enough to solve
the following MINBLOCK problem efficiently.
\noindent
MINBLOCK: Given $\{\alpha,\beta\}\subseteq \Omega$, find the minimum
block containing $\alpha$ and $\beta$. \\
The algorithm is very clever and neat. Define an undirected graph $X =
(V,E)$ such that $V = \Omega$ and $E = \{(\alpha, \beta)\}^G =
\setdef{(\alpha^g,\beta^g)}{g\in G}$.
\begin{claim}
The connected component $C$ containing $\alpha$ is the minimum block.
\end{claim}
\begin{proof}
Note that $G\leq Aut(X)$ and also $G$ is transitive on
$\Omega$. Hence connected components have to move as a whole, and
thus connected components are blocks and hence $C$ is a block.
Suppose $C$ was not minimal, let $C_1 \subsetneq C$ be a block.
Since the containment is strict, there exists an edge $(\gamma,
\delta) = (\alpha^g, \beta^g)$ such that $\gamma \in C_1$ and
$\delta \in C \setminus C_1$. Now $\gamma \in C_1^g \cap C_1$ but
$\delta \in C_1^g \setminus C_1$ which contradicts that $C_1$ is a
block. Hence $C$ has to be the minimal block containing $\alpha$ and
$\beta$.
\end{proof}
One can now run over all $\beta$ for a given $\alpha$ to find solve
the non-trivial block problem.
\section{Membership Testing}
{\bf Problem:} Given $\langle A \rangle = G \leq Sym(\Omega)$ and
$g\in Sym(\Omega)$, check if $g\in G$. \medskip
This problem clearly reduces to the problem of computing the order of
the group given by a set of generators (to check for membership of
$g$, throw $g$ into the generating set and check if the order
changes). We are looking for a divide and conquer approach to solve
membership.
A promising avenue is the orbit-stabilizer formula $|G| =
|G_\alpha||\alpha^G|$. We know to compute $\alpha^G$ efficiently, and
hence can recurse on the smaller subgroup $G_\alpha$. But how do we
get hold of a small generating set for $G_\alpha$? The following
marvellous idea of Schreir gives the answer.
\begin{theorem}[Schreir]
Let $\langle A \rangle = G$ and $H \leq G$. Let $R$ be the set of
distinct coset representatives of $H$, (i.e)
$$
G = \bigsqcup_{r\in R} Hr
$$
Then
$$
B = \setdef{r_1ar_2^{-1}}{a\in A; r_1,r_2\in R}\cap H
$$
generates $H$
\end{theorem}
\begin{proof}
For any given $r_1\in R$ and $a\in A$, there exists a unique $r_2$
such that $r_1ar_2^{-1} \in H$ (because $Hr_1a = Hr_2$).
\begin{eqnarray*}
RA & \subseteq & BR \ \ \ (r_1a = (r_1ar_2^{-1})r_2)\\
& \subseteq & \langle B \rangle R\\
\implies RAA & \subseteq & \langle B\rangle RA\\
& \subseteq & \langle B\rangle \langle B \rangle R = \langle B
\rangle R\\
\therefore \forall t\geq 0,\ RA^t &\subseteq & \langle B \rangle
R\\
\implies G & = & \langle B \rangle R
\end{eqnarray*}
Now since $G$ can be partitioned into cosets of $H$ and there are
distinct representatives of every coset in $R$, unless $\langle B
\rangle = H$, $\langle B \rangle R$ cannot cover the group. Hence
$\langle B \rangle = H$.
\end{proof}
One could then look at $H = G_\alpha$ and the coset representatives
are $g_\beta:\alpha\mapsto\beta$ and can be found in the orbit
computation itself. But this still doesn't quite solve the problem
since the size of the generating set is growing rapidly ($|B| =
|R||H|$) and would get to exponential size in $n$ steps.
This can also be tackled in a very clever way. We shall just see the
sketch here, the details will be worked out next class.
\noindent
{\bf Idea: } For any two elements $\pi$ and $\psi$ in $B$, if $1^\pi
= 1^\psi$ (both $\pi$ and $\psi$ map $1$ to the same element), replace
$\{\pi, \psi\}$ by $\{\pi, \pi^{-1}\psi\}$. This would then ensure
that the two elements map $1$ to different images now. Repeating this
replacement process, we can ensure that the elements of $B$ are never
larger than $n^2$. \medskip
The details shall be worked out in the next lecture, the interested
reader could take this as an easy exercise though.
\end{document}