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\begin{center}
{\large\textbf{Dartmouth College}\\}
Mathematics 101\\\medskip Homework 5 \small{(due Thursday, Oct 29)}
\end{center}
\begin{enumerate}

\item Semidirect products.  We showed in class that $Aut(\Z_n) \cong
  \Z_n^\times$.

  \begin{enumerate}
  \item Suppose that $H_1$, $H_2$ and $K$ are groups, $\sigma: H_1 \to
    H_2$ is an isomorphism, and \newline $\psi:H_2 \to Aut(K)$ a
    homomorphism, so that $\varphi = \psi\circ \sigma: H_1 \to Aut(K)$
    is also a homomorphism.  Show that $K \rtimes_\varphi H_1 \cong K
    \rtimes_\psi H_2$.
  
  \item Suppose that $H$ and $K$ are groups and $\varphi, \psi:H \to
    Aut(K)$ are monomorphisms with the same image in $Aut(K)$.  Show
    that there exists a $\sigma \in Aut(H)$ such that $\psi =
    \varphi\circ \sigma$.

  \item Suppose that $H$ and $K$ are groups, $\varphi, \psi:H \to
    Aut(K)$ are monomorphisms, and $Aut(K)$ is finite and cyclic.
    Show that $\varphi$ and $\psi$ have the same image in $Aut(K)$.

  \item Let $p < q$ be primes with $p \mid (q-1)$. Let $H$ and $K$ be
    cyclic groups of order $p$ and $q$ respectively.  Let $\varphi,
    \psi: H \to Aut(K)$ be nontrivial homomorphisms. Observing that
    $Aut(K)$ is cyclic, show that $K\rtimes_\varphi H \cong
    K\rtimes_\psi H$.

  \item Let $p < q$ be primes. Show that up to isomorphism, there are
    at most two groups of order $pq$.
  \end{enumerate}



\item Let $H$ be a group, and by $H^n$ denote group which is the
  external direct product of $H$ with itself $n$ times.  Show that the
  symmetric group $S_n$ acts on $H^n$ via $(\sigma, (h_1, \dots, h_n))
  \mapsto (h_{\sigma^{-1}(1)}, \dots, h_{\sigma^{-1}(n)})$.  Note that
  the obvious map $(\sigma, (h_1, \dots, h_n)) \mapsto (h_{\sigma(1)},
  \dots, h_{\sigma(n)})$ is a right action, that is you need the
  inverses so that the permutation representation is a homomorphism.

  \textit{Hint:} This is a bit subtle with notation; you may find the
  following observation useful.  In set theory, $Y^X$ denotes the set
  of functions $f: X\to Y$, so one can interpret $H^n$ as $H^X$ where
  $X = \{1, 2, \dots, n\}$.  Now show that the natural action of $S_n$
  on $X$ induces an action of $S_n$ on $H^X \cong H^n$ as suggested
  above.

\item Let $A$, $B$ be groups and let $|B|=n$.  We know that $B$ acting
  on itself by left translation induces an injective homomorphism
  $\rho:B \to S_n$; this permutation representation is usually called
  the \textit{left regular representation}.  As we saw in the previous
  problem, $S_n$ acts on $A^n$ with permutation representation
  $\varphi: S_n \to Aut(A^n)$.  The composition of $\varphi$ and
  $\rho$, is a homomorphism $\varphi\circ \rho: B \to Aut(A^n)$.  The
  \textit{wreath product} of $A$ by $B$, denoted $A \wr B$, is defined
  to be the semidirect product $A\wr B = A^n
  \rtimes_{\varphi\circ\rho} B$.

  Show (obvious) that the order of $A\wr B$ is $|A|^{|B|}\cdot |B|$.
  From this it is clear that $\Z_2\wr \Z_2$ is a group of order 8.
  Determine its isomorphism class in part by writing out all eight
  elements and finding their orders.

\item The point of this exercise is to show that for $p$ a prime, any
  Sylow $p$-subgroup of $S_{p^2}$ is isomorphic to $\Z_p \wr \Z_p$.
  \begin{enumerate}
  \item First compute the cardinality of a Sylow $p$-subgroup of
    $S_{p^2}$ (without assuming the isomorphism).

  \item The following is outline of a proof is from Rotman's
    \textit{Theory of Groups}, where a more general result is
    established.  Let $B_0 = \{1, 2, \dots, p\}$ and let $B_i = B_0 +
    ip$, so that the set $\{1, 2, \dots, p^2\}$ is the disjoint union
    $B_0\cup \cdots \cup B_{p-1}$.  Consider the permutation $\sigma
    \in S_{p^2}$ which for $b_0 \in B_0$ takes
    \[
    \sigma(b_0 + ip) =
    \begin{cases}
      b_0 + (i+1)p& \textrm{if $i < p-1$,}\\b_0&\textrm{if $i = p-1$}.
    \end{cases}
    \]

    Verify that $B_{i+1} = \sigma(B_i)$ (with the subscripts read
    modulo $p$), and that $\sigma$ has order $p$.

  \item Let $H = H_0$ denote a Sylow $p$-subgroup of $S_p$; observe
    that it is cyclic of order $p$.  Viewing $H_0$ also a subgroup of
    $S_{p^2}$, let $H_1$ be the image of $H_0$ under the action of
    $\sigma$; so for example the cycle $(1\, 2\, 3 \dots p)$ would go
    to $(p+1\ p+2\ \dots 2p)$.  Now let $H_{i+1} = \sigma(H_i)$.  Show
    that $S_{p^2}$ contains a subgroup $K = H_0\cdots H_{p-1} \cong
    H_0 \times \cdots \times H_{p-1}\cong H^{p}$. \textit{Hint:
      Disjoint permutations commute.}

  \item Show that $K \cap \langle \sigma \rangle = \{e\}$.

  \item Rotman now says to show that for $(\tau_0, \tau_1,\dots,
    \tau_{p-1}) \in H_0 \times \cdots \times H_{p-1}$, we have
    \newline $\sigma (\tau_0, \tau_1,\dots, \tau_{p-1}) \sigma^{-1} =
    (\tau_1, \tau_2,\dots, \tau_{p-1}, \tau_0)$.

    This is actually pretty subtle. You are trying to lead up to
    identifying $\langle K, \sigma\rangle$ as a wreath product and so
    you want to observe the standard action on $H^p$.  The subtlety is
    that $H^p \cong H_0\times \cdots \times H_{p-1}$ and you need to
    chase through the identifications to verify his claim.

  \item Conclude that $\langle K, \sigma\rangle \cong H \wr \langle
    \sigma \rangle \cong \Z_p \wr \Z_p$
  \end{enumerate}


\end{enumerate}


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