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\begin{document}
\lecture{14}{12 October 2001}{Igor Pak}{D. Jacob Wildstrom}

\section*{Random Walks on Nilpotent Groups}
\begin{example}
  Let $$G=\left\{\begin{pmatrix}1&\ast&\cdots&\ast\\
                           0&1&\cdots&\ast\\
                           \vdots&\vdots&\ddots&\vdots\\
                           0&0&\cdots&1\end{pmatrix}\:\ast\in\mathbb F_p\right\},$$
  that is to say, the group of upper triangular matrices
  with ones on the diagonal. Let the
  generator set $S$ consist of the {\em elementary transvections} --
  that is, matrices equivalent to the identity matrix except that the
  $(i,j)$-th entry is $a$, for $1<i<j<n$, $a\in\mathbb F_p$. For purposes
  of simplicity we regard each matrix defined by an $(i,j,a)$ triple
  as distinct, even those which are identical (for instance, all
  matrices with $a=0$ are identical). With this enumeration, we may
  easily find that $|S|=\binom n2p$, and that $|G|=p^{\binom n2}$. Now
  let us design a random process in the usual fashion: let $X_0=I$,
  and $X_{t+1}=X_tE_{ij}(a)$ where $E_{ij}(a)$ is a random variable
  uniformly distributed in $S$.

  We may more intuitively represent $XE_{ij}(a)$ as identical to $X$,
  but with the $i$th row increased by the $j$th row multiplied by $a$.

\end{example}

\begin{theorem}
  The mixing time of a random walk $\{X_t\}$ on
  $\Gamma(G,S)$ is $O(n^2\log n)$.
\end{theorem}
\begin{proof}
  Let us define the stopping time $\varkappa$ by the following rule: we
  wait until all pairs $i,j$ have been used in the walk (we place no
  restrictions on the value of $a$). By the Coupon Collector's
  problem, $E(\varkappa)\approx\binom n2\log\binom n2=O(n^2\log n)$, and
  by the following lemma, $\tau_3\leq E(\varkappa)$.
\end{proof}

\begin{lemma}
  $\varkappa$ is strong uniform.
\end{lemma}

Before we address random walks further though, we'll want to discuss
the non-probabilistic properties of nilpotent groups.

\section*{General Group Theory}

First, let's review some facts we all learned in kindergarten. For a
finite group $G$ and subgroup $H$ normal in $G$, $G/H$ is the quotient
group; the multiplicative group of cosets of $H$. Note that
$|G/H|=\frac{|G|}{|H|}=[G:H]$.

Another important concept is the commutator group: for $H\triangleleft G$, we define $[G,H]$ as the group generated by all products of the form $ghg^{-1}h^{-1}$, for $g\in G$ and $h\in H$.

\begin{proposition}
  If $H\triangleleft G$, then $[G,H]\triangleleft G$. Moreover,
  $G/[G,G]$ is
  abelian.
\end{proposition}
\begin{proof}

  Since the conjugate of a product is the product of conjugates, it is
  only necessary to show that the generators of $[G,H]$, under
  conjugation by elements of $G$, remain in $[G,H]$, and from this the
  normality of the entire group $[G,H]$ follows. Let $a\in G$, and
  $ghg^{-1}h^{-1}$ be a generator of $[G,H]$. Then
  $$(ghg^{-1}h^{-1})^a=g^ah^a(g^a)^{-1}(h^a)^{-1}$$
  Since $H\triangleleft G$, $g^a\in G$ and $h^a\in H$, so
  $g^ah^a(g^a)^{-1}(h^a)^{-1}$ is in the commutator group [G,H].

  From here, proving that $G/[G,G]$ is abelian is quite
  straightforward: let $a,x\in G$, and let $[a]$ and $[x]$ be the
  associated elements of $G/[G,G]$. Then
  $$[x][a]=[xa]=[xa(a^{-1}x^{-1}ax)]=[ax]=[a][x]$$
\end{proof}

Finally, we shall discuss application of these properties to nilpotent
groups.

\begin{definition}
  The {\em lower central series} of a group $G$ is the chain of groups
  $G_0\triangleright G_1\triangleright G_2\triangleright\cdots$
  defined by $G_0=G$ and $G_{i+1}=[G,G_i]$.
  Group $G$ is called {\em nilpotent} if
  some $G_\ell$ in the lower central series of $G$ is trivial.
\end{definition}

Let $H_i=G_{i-1}/G_i$. Each $H_i$ is abelian since
$G_i=[G,G_{i-1}]\supset[G_{i-1},G_{i-1}]$, thus $G_{i-1}/G_i\subset
G_{i-1}/[G_{i-1},G_{i-1}]$, which was shown earlier to be abelian.

\begin{definition}
  For prime $p$, a finite group $G$ is called a {\em $p$-group}
  if  $|G|=p^m$ for some integer~$m$.
\end{definition}

\begin{theorem}
  If $G$ is a finite $p$-group, then $G$ is nilpotent.
\end{theorem}

\begin{example}
  The group of upper triangular matrices $U(n,p)$ of $\mathbb F_p$ is
  a $p$-group, hence nilpotent. Calculation of the lower central series
  yields:
  $$G_1=\begin{pmatrix}
          1&0&\ast&\ast&\cdots&\ast\\
           &1&0&\ast&\cdots&\ast\\
           &&1&0&\cdots&\ast\\
           &&&\ddots&\ddots&\vdots\\
  %         &&&1&0&1 \\
           &&&&1&0 \\
           &&&&&1
        \end{pmatrix} \, , \qquad
    G_2=\begin{pmatrix}
          1&0&0&\ast&\cdots&\ast\\
           &1&0&0&\cdots&\ast\\
           &&1&0&\cdots&\ast\\
           &&&\ddots&\ddots&\vdots\\
   %        &&&1&0&0 \\
           &&&&1&0 \\
           &&&&&1
        \end{pmatrix} \,,$$
  and so forth, so that $G_{n-1}$ consists only of the identity
  matrix. Interestingly enough, we can deduce from this that
  $H_i\simeq\mathbb Z_p^{n-1}$.
\end{example}

Finally, another probabilistic result:
\begin{lemma}
  Suppose $N\triangleleft G$. Let $H=G/N$, and let
  $\gamma:G\rightarrow H$ be the standard onto map from $g\in G$ to
  the coset $gN\in H$. For any map $\psi:H\rightarrow G$ such that
  $\gamma(\psi(h))=h$ for all $h\in H$, the formula $\psi(h)n$ is
  uniform in $G$ given that $h$ and $n$ are uniform in $H$ and $N$.
\end{lemma}
\begin{proof}
  This is obvious, according to Pak's notes. To go into slightly more
  detail, it's a well-known fact from algebra that the cosets of $N$
  are disjoint, equal size, and cover $G$. For $\psi$ to satisfy the
  given condition, it must map each coset $aN$ to one of its elements.
  SO, the product $\psi(h)n$ is essentially a uniform selection from
  the cosets $aN$, then a uniform selection from the cosets
  elements. The partition and equality conditions of cosets guarantee
  that such a selection process is uniform.
\end{proof}
\end{document}
