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{{\Large\bf MODERN ALGEBRA II}\\
\medskip

{\Large\bf HOMEWORK ASSIGNMENT VIII}\\
\medskip
{\bf (Turn in underlined problems.)}}
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%\vspace{.25in}

\medskip

\noindent
\underline{\large\bf PROBLEMS}
\vspace{.25in}

{\bf 

The purpose of these problems is to determine all groups
$G$ of order $pq$ where $p$ and $q$ are distinct primes.
Assume throughout that $p$ and $q$ are distinct primes
and that $|G| = pq$.

\medskip

%\underline{1.}
1. Show that $G$ has subgroups $P$ and $Q$ such that
$P\cong {\BZ}_p, Q\cong {\BZ}_q, P\cap Q = \{1\}$, and $PQ = G$.

\medskip

% \underline{2.} 
2. Show that either $P\lhd G$ or $Q\lhd G$.
(Hint: Let $X = \{P, P_2, \ldots, P_m\}$ 
be the set of conjugates of $P$ and let 
$G$ act on $X$ by congugation: $\widehat{g}(P_i) =
g^{-1}P_ig$. Explain why $X$ is a transitive $G$--set
and why the stabilizer of the point $P$ contains
the subgroup $P$. Conclude that $P$ has either
1 or $q$ conjugates. If $P$ is not normal, then
it has $q$ conjugates and these conjugates contain
$q(p-1)$ elements of order $p$.
Similarly, if $Q$ is not normal then the conjugates
of $Q$ contain $p(q-1)$ elements of order $P$.
Show that this accounts for too many elements.)

\smallskip

The previous exercise guarantees that 
$G\cong {\BZ}_p\times_{\varphi} {\BZ}_q$ for some 
$\varphi:{\BZ}_p\rightarrow {\rm Aut}({\BZ}_q)^{op}$
or else 
$G\cong {\BZ}_q\times_{\varphi} {\BZ}_p$ for some 
$\varphi:{\BZ}_q\rightarrow {\rm Aut}({\BZ}_p)^{op}$.
(See HW 6.)

\smallskip

% \underline{3.} 
3. Let $r$ be a primitive root modulo $q$.
Show that the function $\pi(x) = rx$ is an
automorphism of ${\BZ}_q$. Explain why every 
automorphism of ${\BZ}_q$ is one of
$\{\pi, \pi^2,\ldots,\pi^{q-2},\pi^{q-1} = {\sf id}\}$. 
Conclude that ${\rm Aut}({\BZ}_q)\cong {\BZ}_{q-1}$.

\medskip

% \underline{4.} 
4. Show that if $p$ does not divide 
$q-1$, then the only homomorphism $\varphi:{\BZ}_p\rightarrow 
{\rm Aut}({\BZ}_q)^{op}$ is the constant function
$\varphi(x) = 1$. Use this 
to explain why ${\BZ}_p\times_{\varphi}{\BZ}_q
= {\BZ}_p\times {\BZ}_q\cong{\BZ}_{pq}$ 
when $p$ does not divide $q-1$.


\medskip

5. Assume that $p < q$. Show that 
${\BZ}_{pq}$ is the only abelian group of order $pq$.
Show that there is a nonabelian group of order $pq$
if and only if $p\;\mid\;q-1$, and when this condition
is met there is exactly one nonabelian group of
order $pq$.

\medskip

For Graduate Credit: 

H. Find a number $j$ for which 
$\langle a, b\;|\;a^{23} = b^{11} = 1, b^{-1}ab = a^j\rangle$
describes the nonabelian group of size $23$.
There is more than one answer to this question.
(Hint: The solution to the last part of
Additional Problem 3 from HW 7 
could be any $r\in \{5, 7,
10, 11, 14, 15, 17, 19, 21\}$. It helps
to know this.)

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