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\begin{document}
\lecture{28}{21 November 2001}{Igor Pak}{Fumei Lam}

\section*{Product Replacement Graphs}

\begin{definition} Let $G$ be a finite group and let $k \geq d(G)$, where $d(G)$
is the minimum number of generators of $G$.  The product replacement graph
$\Gamma_k (G)$ is a graph on $k$-tuples $(g_1, g_2, \ldots g_k) \in G^k$
satisfying $< g_1, g_2, \ldots g_k > = G$.  The edges of $\Gamma_k(G)$
are

$$\{ \overline{g}, \, R_{ij}^{\pm}(\overline{g} ) \, \},$$
$$\{ \overline{g}, \, L_{ij}^{\pm}(\overline{g} ) \, \},$$

\noindent where

$$R_{ij}^{\pm}(g_1, \ldots g_i, \ldots g_j, \ldots g_k) = (g_1, \ldots
g_ig_j^{\pm 1}, \ldots g_j, \ldots g_k) \},$$
$$L_{ij}^{\pm}(g_1, \ldots g_i, \ldots g_j, \ldots g_k) = (g_1, \ldots
g_j^{\pm 1}g_i, \ldots g_j, \ldots g_k) \}.$$

\end{definition}

There are $k(k-1)$ choices for pairs $(i,j)$, and two choices each for $R$
or $L$ and $+$ or $-$.  By allowing vertices in $\Gamma_k(G)$ to contain
loops, this implies $\Gamma_k(G)$ is a $D-$ regular graph with $D =
4k(k-1)$.

{\bf Example.}  For $G = Z_p^m, d(G) = m$, the vertices of
$\Gamma_m(Z_p^m)$ are matrices

$$A = \left[ \begin{array}{cccc}
     a_{11} & a_{12} & \cdots & a_{1m} \\
       a_{21} & a_{22} & \cdots & a_{2m} \\
        \vdots & \vdots & \vdots & \vdots \\
       a_{m1} & a_{m2} & \cdots & a_{mm} \\
\end{array} \right]$$

with $det(A) \not= 0, a_{ij} \in F_p$.

The operations $R_{ij}^{\pm}$ and $L_{ij}^{\pm}$ correspond to
left multiplication by

$$E^{\pm} = \begin{cases} 1 \text{ on the diagonal } \\
              \pm 1 \text{ in entry ij }\\
              0 \text{ otherwise}.\\
        \end{cases}
$$

Note that since the group is abelian, the operations $L$ and $R$
are the same.  So $\Gamma_m(Z_p^m)$ is the Cayley graph
$\Gamma(GL(m,p), \{ E_{ij}(\pm 1)  \} )$.  Since $E^{\pm}$ has
determinant $\pm 1$, $\Gamma_m(Z_p^m)$ has $p-1$ connected
components, each corresponding to different values for the
determinant.

\begin{conjecture}
If $k \geq d(G) + 1$, then $\Gamma_k(G)$ is connected.
\end{conjecture}

The following weaker conjecture is also unknown.

\begin{conjecture}
If $k \geq 3$, then $\Gamma_k(S_n)$ is connected.
\end{conjecture}

\begin{lemma} (Higman) Let $k = 2$.  Then the conjugacy class of $[g_1,
g_2]$  ($< g_1, g_2 > = G$) is invariant on connected components of
$\Gamma_2(G)$.
\end{lemma}

\begin{proof}
For $(g_1, g_2) \in V(\Gamma_2(G))$, $\{ L^{\pm} (g_1, g_2),
R^{\pm} (g_1, g_2) \} = \{ (g_1g_2, g_2), (g_1g_2^{-1}, g_2), (g_2g_1,
g_2), (g_2^{-1}g_1, g_2) \}$.

Then

$$[g_1g_2, g_2] = g_1g_2g_2g_2^{-1}g_1^{-1}g_2^{-1} = [g_1,g_2],$$
$$[g_2g_1, g_2] = g_2g_1g_2g_1^{-1}g_2^{-1}g_2^{-1} = g_2[g_1,g_2]g_2^{-1}
=[g_1,g_2]^{g_2^{-1}},$$
$$[g_1g_2^{-1}, g_2] = g_1g_2^{-1}g_2g_2g_1^{-1}g_2^{-1} =
[g_1,g_2],$$
$$[g_2^{-1}g_1, g_2] = g_2^{-1}g_1g_2g_1^{-1}g_2g_2^{-1} =
g_2^{-1}[g_1,g_2]g_2
=[g_1,g_2]^{g_2}.$$

\end{proof}

{\bf Example.}  Let $G = A_n$ for $n$ odd, $k = 2$ and consider
$a = (1 2 3 \ldots n), b = (1 2 3 \ldots p)$ with $p \not| n$.

Then the commutators lie in different conjugacy classes, implying the
number of connected components of $\Gamma_2(A_n) \rightarrow \infty$ as $n
\rightarrow \infty$.

\begin{theorem}
Let $m(G)$ denote the maximum size of a nonredundant generating set.  For
$k \geq d(G) + m(G)$, $\Gamma_k(G)$ is connected.
\end{theorem}

In order to prove the theorem, we first define graph
$\widetilde{\Gamma}_k(G)$  as the graph $\Gamma_k(G)$ with additional
edges defined by the operators

$$I_m(g_1, g_2, \ldots g_m, \ldots g_k) = (g_1, g_2, \ldots g_m^{-1},
\ldots g_k)$$
$$\pi_{ij}(g_1, g_2, \ldots g_i, \ldots g_j, \ldots g_k) = (g_1, g_2,
\ldots g_j, \ldots g_i, \ldots g_k).$$

Then we have the following lemma.

\begin{lemma}
The number of connected components of $\Gamma_k(G)$ is less than or equal
to the number of connected components of $\widetilde{\Gamma}_k(G)$.
\end{lemma}

\begin{proof}
Define the operation $T_{ij}(g_1, g_2, \ldots g_i, \ldots g_j,
\ldots g_k) = (g_1, g_2, \ldots g_j^{-1}, \ldots g_i, \ldots
g_k)$, i.e., $T_{ij}$ switches entries $g_i$ and $g_j$ and
inverts $g_j$.  Note that $T_{ij} =
L_{ij}^{-}L_{ji}^{+}R_{ij}^{-}$ and

$$T_{ij}^2 (g_1, g_2, \ldots g_i, \ldots g_j, \ldots g_k) = (g_1, g_2, \ldots
g_i^{-1} \ldots g_j^{-1} \ldots g_k).$$

Now note that since $\Gamma_k(G)$ is a subgraph of
$\widetilde{\Gamma}_k(G)$, this implies every connected component
of $\widetilde{\Gamma}_k(G)$ splits into at most 2 components in
$\Gamma_k(G)$.

\end{proof}

As a corollary, if $\widetilde{\Gamma}_k(G)$ is connected for $k
\geq d(G) + 1$, then $\Gamma_k(G)$ is connected.

\newpage

Since $m(G) \geq 1$ for all groups, we need only consider
$\widetilde{\Gamma}_k(G)$ to prove the theorem.

\begin{theorem} For $k \geq d(G) + m(G)$, $\widetilde{\Gamma}_k(G)$ is
connected. \end{theorem}

\begin{proof}

By definition of $m = m(G)$, any element $(g_1, g_2, \ldots
g_k) \in \Gamma_k(G)$ contains a generating subset of $m$ elements
$g_{i_1}, g_{i_2}, \ldots g_{i_m}$.  Use the operators $\pi_{ij}$ to
obtain

$$(g_1, g_2, \ldots g_k) \rightarrow (g_{i_1}, g_{i_2}, \ldots g_{i_m},
\ldots )$$

where the remaining $k - m$ elements are those not in
$\{ g_{i_1}, g_{i_2}, \ldots g_{i_m} \} $.  Now since $g_{i_1}, g_{i_2},
\ldots g_{i_m}$ form a generating set, we can use the $L^{\pm}$ and
$R^{\pm}$ operations to obtain

\begin{eqnarray*}
(g_{i_1}, g_{i_2}, \ldots g_{i_m}, \ldots) & \rightarrow &
(g_{i_1}, g_{i_2}, \ldots g_{i_m}, 1, 1, \ldots 1 ) \\
 & \rightarrow & ( g_{i_1}, g_{i_2}, \ldots g_{i_m}, h_1, h_2, \ldots
h_{k-m} ),
\end{eqnarray*}

where $h_1, h_2, \ldots h_{k-m}$ is a generating set of $G$ (this is
possible, since $k-m \geq d$).  Then we again use the $L^{\pm}, R^{\pm}$
operators to obtain

$$ ( g_{i_1}, g_{i_2}, \ldots g_{i_m}, h_1, h_2, \ldots
h_{k-m}) \rightarrow (1, 1, \ldots 1, h_1, h_2, \ldots h_{k-m}).$$

Therefore, every element in $\widetilde{\Gamma}_k(G)$ is connected
to the element $(1, 1, \ldots 1, h_1, h_2, \ldots h_{k-m})$, implying
$\widetilde{\Gamma}_k(G)$ is connected.

\end{proof}

Now Theorem 5 follows immediately from Lemma 6 and Theorem 7. We
also have the following corollary.

\begin{corollary}
For $k \geq 2 \lfloor \log_2 \vert G \vert \rfloor, \Gamma_k(G)$ is
connected.
\end{corollary}

The following theorem shows that $\Gamma_3(A_n)$  contains very large
connected components.

\begin{theorem}
There exists $\Gamma' \subset \Gamma_k(A_n)$ such that $\Gamma'$ is
connected for all $k \geq 3$ and

$$\frac{\vert \Gamma' \vert}{\vert \Gamma_k(A_n) \vert} \rightarrow 1
\text{ as } n \rightarrow \infty.$$

\end{theorem}

\begin{proof}
For $k = 3$, pick $g_1, g_2, g_3, h_1, h_2, h_3 \in A_n$ uniformly and
independently.  We will show that with high probability, the elements
$(g_1, g_2, g_3), (h_1, h_2, h_3) \in \Gamma_3(A_n)$ are connected.
Since $< g_1, g_2 > = A_n$ with high probability, we can use $L^{\pm},
R^{\pm}$ operations to obtain

$$(g_1, g_2, g_3) \rightarrow (g_1, g_2, h_3).$$

Similarly, since $h_2$ and $h_3$ were chosen uniformly, $< g_1, h_3 > =
<h_2, h_3> = A_n$ with high probability, so we have

$$(g_1, g_2, h_3) \rightarrow (g_1, h_2, h_3) \rightarrow (h_1, h_2,
h_3).$$

Since the probability two random elements generate $A_n$ is at
least $1 - \frac{1}{n}$, the probability two random elements are
connected is at least $1 - \frac{1}{3n}$ and the theorem follows.

\end{proof}

\end{document}
