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\begin{center}{\bf M901, Assignment 5: Due Friday, October 28, 2011}\end{center}

{\it Instructions}: Do any three problems.

Background: Let $N$ and $H$ be groups and let $\theta:H\to\operatorname{Aut}(N)$ be a homomorphism.
We define $N\rtimes_\theta H$ to be the set $N\times H$ with the composition law
$(n_1,h_1)(n_2,h_2)=(n_1(\theta(h_1))(n_2),h_1,h_2)$. It is now easy to check that
$(n,h)(1_N,1_H)=(n((\theta(h))(1_N)),h1_H)=(n1_N,h1_H)=(n,h)$ and similarly that 
$(1_N,1_H)(n,h)=(1_N((\theta(1_H))(n)),1_Hh)=(1_Nn,1_Hh)=(n,h)$, so $(1_N,1_H)$ is an identity. Moreover,
$$((n_1,h_1)(n_2,h_2))(n_3,h_3)=(n_1((\theta(h_1))(n_2)),h_1h_2)(n_3,h_3)=(n_1((\theta(h_1))(n_2))((\theta(h_1h_2))(n_3)),h_1h_2h_3)$$
while 
$$(n_1,h_1)((n_2,h_2)(n_3,h_3))=(n_1,h_1)(n_2((\theta(h_2))(n_3)),h_2h_3)=(n_1((\theta(h_1))(n_2((\theta(h_2))(n_3)))), h_1h_2h_3),$$
but these are the same (and hence the composition law is associative) since
$$(\theta(h_1))(n_2((\theta(h_2))(n_3)))=((\theta(h_1))(n_2))((\theta(h_1))((\theta(h_2))(n_3)))=
((\theta(h_1))(n_2))((\theta(h_1h_2))(n_3)).$$

Note that $(n,h)=(n,1_H)(1_N,h)$. It is convenient to identify $N$ with $N\times (1_H)$ and $H$ with $(1_N)\times H$;
i.e., $n\in N$, regarded as an element of $N\rtimes_\theta H$, is $(n,1_H)$, and $h\in H$,
regarded as an element of $N\rtimes_\theta H$, is $(1_N,h)$.
With these conventions $nh$ denotes $(n,1_H)(1_N,h)$; i.e., $nh=(n,h)$. If we do this, then we should regard
$hn$ as denoting $(1_N,h)(n,1_H)=((\theta(h))(n),h)$, which we can denote as $(\theta(h)(n))h$. 
It is convenient to denote $(\theta(h))(n)$ by $n^h$. Combining the two notations gives $hn=n^hh$.
Let's check associativity using these conventions: 
$$(n_1h_1)((n_2h_2)(n_3h_3))=
(n_1h_1)((n_2n_3^{h_2})(h_2h_3))=
(n_1(n_2n_3^{h_2})^{h_1})(h_1h_2h_3)=
(n_1n_2^{h_1}n_3^{h_1h_2})(h_1h_2h_3)$$
and
$$((n_1h_1)(n_2h_2))(n_3h_3)=
((n_1n_2^{h_1})(h_1h_2))n_3h_3=
(n_1n_2^{h_1}n_3^{h_1h_2})(h_1h_2h_3).$$
Note that under these identifications we have $1_N=1_H=1_{N\rtimes_\theta H}$.


\begin{itemize}

\item[(1)] 
\begin{itemize}
\item[(a)] Show that $(n,h)$ has an inverse in $N\rtimes_\theta H$ by writing it down explicitly and checking.
Conclude that $N\rtimes_\theta H$ is a group.
\item[(b)] Using the conventions above, show that $n^h=hnh^{-1}$. (Thus $\theta(h)$ applied to $N$
in $N\rtimes_\theta H$ is just conjugation by $h$.)
\item[(c)] Show that $N$ (regarded as $N\times(1_H)$) is a normal subgroup of $N\rtimes_\theta H$.
\end{itemize}
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\item[(2)] Let $N$ and $H$ be groups and let $\theta:H\to\operatorname{Aut}(N)$ be a homomorphism.
Show that $N\rtimes_\theta H$ is abelian if and only if $\theta$ is trivial and $N$ and $H$ are abelian.

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\item[(3)] Let $N$ and $H$ be groups and let $\theta:H\to\operatorname{Aut}(N)$ be a homomorphism.
Let $\phi:N\times H\to N\rtimes_\theta H$ be the identity on elements.
Show that the following are equivalent.
\begin{itemize}
\item[(a)] $\phi$ is an isomorphism;
\item[(b)] $\phi$ is a homomorphism; and
\item[(c)] $\theta$ is trivial.
\end{itemize}

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\item[(4)] Let $G$ be a group. Let $\operatorname{Inn}(G)$ be the group of \emph{inner automorphims}, i.e., 
the subgroup of $\operatorname{Aut}(G)$ of automorphisms of the form
$\alpha_g$ for $g\in G$, defined 
for any $h\in G$ by $\alpha_g(h)=ghg^{-1}$.
Show that $\operatorname{Inn}(G)\lhd\operatorname{Aut}(G)$.
(Aside: the quotient $\operatorname{Aut}(G)/\operatorname{Inn}(G)$ is known
as the outer automorphism group, sometimes denoted $\operatorname{Out}(G)$, 
while an automorphism which is not inner is
called an outer automorphism. Unfortunately, the elements of $\operatorname{Out}(G)$
are not usually themselves automorphisms, but rather cosets of
the group of inner automorphisms.)
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\item[(5)] Construct a non-abelian semi-direct product $N\rtimes_\theta H$ of order 27, where neither
$N$ nor $H$ has order 27.
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\item[(6)] Let $G$ be an abelian group and let $H$ be a subgroup of $G$.
Let $q:G\to G/H$ be the quotient.
\begin{itemize}
\item[(a)] If there is a 
homomorphism $s:G/H\to G$ such that the composition
$qs$ is the identity, show that $\phi:H\times (G/H)\to G$ defined as 
$\phi(h,x)=h+s(x)$ is an isomorphism.
\item[(b)] Assume that every element of $G$ has order either 1 or 2.
Show that there is an injective 
homomorphism $s:G/H\to G$ such that the composition
$qs$ is the identity. [Hint: Apply Zorn's Lemma.]
\end{itemize}
(Aside: In the situation of part (a) we have a short exact sequence
$$0\to H\to G\to G/H\to 0$$
of abelian groups and we say that $s$ is a splitting or that
the short exact sequence is split when there is a homomorphism $s$ with $qs$
being the identity. Given a split short exact sequence, the middle
is isomorphic to the direct product of the ends.)

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\item[(7)] Let $G$ be a group. Show 
that $|\operatorname{Aut}(G)|> 1$ if and only if $|G|>2$.

\end{itemize}
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