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\begin{document}
\lecture{30}{28 November 2001}{Igor Pak}{Michael Korn}

\section*{Bias in the Product Replacement Algorithm}
Here is the algorithm:

The input to the algorithm is a $k$-tuple $\overline{g} = (g_1, g_2, \ldots, g_r, 1, \ldots, 1)$,
where the elements $g_1, \ldots, g_r$ generate the group $G$.

We then run a random walk on $\Gamma_k(G)$ starting at $\overline{g}$ for $L$ steps, putting us at the point
$\overline{g'}$.  We choose $i$ randomly from ${1, \ldots, k}$, and output the group element $g'_i$.

This algorithm is supposed to generate random group elements.

Here are some questions which can be asked about this algorithm:

Q1:  Is $\Gamma_k(G)$ connected?

Q2:  How do we choose the values for $k$ and $L$?

Q3:  Is there bias in the output?  (Are all group elements equally represented in generating $k$-tuples?)

In this lecture we will try to answer question 3.

\begin{definition}
Suppose $G$ is a finite group, and $k \geq d(G)$.  (Recall that $d(G)$ is the minimum number of generators necessary
to generate $G$.)

Let $Q$ be the probability distribution of the first component of $(g_1, \ldots, g_k)$, where
$(g_1, \ldots, g_k)$ is selected uniformly at random from among all $k$-tuples which generate $G$.
So $Q(a)$ is the probability that $g_1 = a$.
\end{definition}

\begin{proposition}
Let $\phi_k(G)$ be the probability that a random $k$-tuple $(g_1, \ldots, g_k)$ generates $G$.

If $\phi_{k-1}(G) \geq 1/2$, then $sep(Q) \leq 1/2$.
\end{proposition}

\begin{proof}

We need to show that for all $a \in G$, $Prob(g_1 = a) \geq \frac{1}{2|G|}$, where $(g_1, \ldots, g_k)$
is a random generating $k$-tuple.  This probability is equal to the number of generating $k$-tuples of the
form $(a, g_2, \ldots, g_k)$, divided by the total number of generating $k$-tuples.  The total number of
generating $k$-tuples is at most $|G|^k$.  Since $\phi_{k-1}(G) \geq 1/2$, the $(k-1)$-tuple $(g_2, \ldots, g_k)$
generates $G$ at least half the time.  So $(a, g_2, \ldots, g_k)$ is a generating $k$-tuple at least half the time,
for any $a$.  So there are at least $\frac{|G|^{k-1}}{2}$ generating $k$-tuples which have first element $a$.  So
the probability is at least $\frac{1}{2|G|}$.
\end{proof}

Question:  Are there finite groups $G$ with very small $\phi_k(G)$ for $k \geq d(G)$?

The answer will turn out to be yes.  Let $G_n$ be the group $(A_n)^{n!/8}$.  Then $d(G_n) = 2$ for $n$ large enough.
This fact follows from the following theorem.

\begin{theorem} (P. Hall, 1938)

Let $H$ be a nonabelian simple group.  Let $\alpha_k(H) = \max\{m: d(H^m)=k\}$.  Then $\alpha_k(H)$ is the number of $Aut(H)$
orbits of action on $\Gamma_k(H)$.
\end{theorem}

Let us see why this implies the earlier fact.  We let $H = A_n$ and let $k = 2$.  For $n>6$, it is a fact that
$Aut(A_n)=S_n$.  (This is not true for $n=6$, but this is not for normal people to understand why.)  We will assume
that $\phi_2(A_n) \geq 1/2$ (so two random elements of $A_n$ generate $A_n$ at least half the time).  Thus there are
at least $\frac{1}{2} (\frac{n!}{2})^2$ vertices in $\Gamma_2(A_n)$.  Since $Aut(A_n) = S_n$, the size of an orbit
is $n!$, so the number of orbits is at least $\frac{1}{2} (\frac{n!}{2})^2 \frac{1}{n!} = \frac{n!}{8}$.  So
$\alpha_2(A_n) \geq \frac{n!}{8}$, so $d((A_n)^{n!/8}) = 2$, which proves the fact from above.

We will now give a proof of Hall's Theorem.

\begin{proof}

Let $G = H^m$.  Take $\langle g_1, \ldots, g_k \rangle = G$, and let $g_i = (h_1^{(i)}, h_2^{(i)}, \ldots, h_m^{(i)}) \in G$,
where $h_j^{(i)} \in H$.  Let us write these elements in a $k$-by-$m$ array as shown:

$g_1 = \ \ \ \ \ \ h_1^{(1)}, h_2^{(1)}, \ldots, h_j^{(1)}, \ldots, h_m^{(1)}$

$g_2 = \ \ \ \ \ \ h_1^{(2)}, h_2^{(2)}, \ldots, h_j^{(2)}, \ldots, h_m^{(2)}$

$\ \ \vdots$

$g_k = \ \ \ \ \ \ h_1^{(k)}, h_2^{(k)}, \ldots, h_j^{(k)}, \ldots, h_m^{(k)}$

Now look at the columns of this array.  For all $j$, we must have $\langle h_j^{(1)}, h_j^{(2)}, \ldots, h_j^{(k)} \rangle = H$.

\begin{claim}
$\langle g_1, \ldots, g_k \rangle = G$ iff $(h_j^{(1)}, \ldots, h_j^{(k)})$ are generating $k$-tuples in different
$Aut(H)$ orbits.
\end{claim}

Proving this claim is enough to prove the theorem.

The ``only if'' direction is obvious; if two such $k$-tuples were in the same orbit, then it would be impossible for
$(g_1, \ldots, g_k)$ to generate all of $H^m$, since the two columns would always be bound by the isomorphism between them.

For the ``if'' direction, assume the columns of the array are generating $k$-tuples which are in different orbits.   Let
$B = \langle g_1, \ldots, g_k \rangle$, and suppose $B$ does not equal $G$.  We will use an inductive argument.  So assume
that for a $k$-by-$(m-1)$ array, if the columns are generating $k$-tuples in different orbits, then the rows generate all
of $G$.  In our situation, this means that the projection of $B$ onto the first $m-1$ coordinates is onto.  (In other words,
for any choice of the first $m-1$ coordinates, there is an element of $B$ which attains those values, though we can't say
what its last coordinate will be.)  Of course, there is nothing special about the first $m-1$ coordinates; this statement
holds for any collection of $m-1$ coordinates.

Now consider the subset $C \subset B$ consisting of points whose first $m-1$ coordinates are all equal to 1
(the identity element).  This is a normal subgroup of $H$, hence it is either $H$ itself, or 1, since $H$ is simple.
Suppose $C = H$.  Then $B$ would have to equal $G$, since we can find an element of $B$ which sets the first $m-1$
coordinates to whatever we want, and then multiplying this by an appropriate member of $C$ will yield any element
of $G$ at all.  So we must assume $C = 1$.  Again, there is nothing special about the last coordinate.  So any element
of $B$ which has the value 1 for $m-1$ of its coordinates must have value 1 in the remaining coordinate as well.

Recall that we assumed $H$ is nonabelian.  Hence there are elements $x$ and $y$ with $xyx^{-1}y^{-1} \ne 1$.  Since we can
set any $m-1$ coordinates any way we like, it follows that $(x, 1, \ldots, 1, z) \in B$, for some $z$.  Similarly, we have
$(y, 1, \ldots, 1, w, 1) \in B$, for some $w$.  Multiplying these gives $(xy, 1, \ldots, 1, w, z) \in B$ and
$(yx, 1, \ldots, 1, w, z) \in B$.  Dividing these last two gives $(xyx^{-1}y^{-1}, 1, \ldots, 1, 1) \in B$.  But this
contradicts the result of the previous paragraph.

This completes the proof of Hall's Theorem.

\end{proof}

Now back to the situation with $A_n$.  As it turns out, for $A_5$ we have $\alpha_2(A_5) = 19$, which is greater
than $\frac{5!}{8} = 15$, as we claimed that it should be for $n$ large enough.

\begin{claim}
$\phi_k((A_n)^{n!/8}) \rightarrow 0$ very rapidly as $n \rightarrow \infty$.
\end{claim}

\begin{proof}

Recall that
$$Prob(\langle \sigma_1, \sigma_2, \ldots, \sigma_k \rangle \neq A_n) > \frac{1}{n^k}$$
(This is true since each permutation will fix the point 1 with probability $1/n$, hence with probability
$1/n^k$ all $k$ permutations will fix the point 1.)  So
$$Prob \left(\langle g_1, \ldots, g_k \rangle = (A_n)^{n!/8} \right) \leq \left(1-\frac{1}{n^k}\right)^{n!/8} \leq e^{-{n!}/{8n^k}}$$
which is very small for $n$ large.

\end{proof}

\end{document}
