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\begin{document}

\noindent
{\large Semidirect Products.}

\smallskip

Let $K$ be an abelian group.
For each exact sequence 
$\displaystyle 
0\rightarrow K \stackrel{\alpha}{\rightarrow} G
\stackrel{\beta}{\rightarrow} Q\rightarrow 1$
there is an induced homomorphism 
$\theta:Q\rightarrow {\rm Aut}(K)$, defined 
by 
\begin{eqnarray} \label{eqn}
\theta(q)(k) =
\alpha^{-1}\left(\left[\beta^{-1}(q)\right]\alpha(k)
\left[\beta^{-1}(q)\right]^{-1}\right).
\end{eqnarray}
We say that the extension {\em realizes $\theta$}.

Th intimidating-looking formula (\ref{eqn}) has a simple
explanation. For each $q\in Q$ we want to 
define an automorphism $\theta(q)\in {\rm Aut}(K)$.
To describe this automorphism we need to explain
how to determine
$\theta(q)(k)$ for each $k\in K$.
The idea is to identify $K$ and $Q$ with
subsets $\alpha(K), \beta^{-1}(Q)\subseteq G$,
and then (since everything is taking place
in $G$) let $\theta$ describe the action of
$\beta^{-1}(Q)$ on $\alpha(K)$ by conjugation.
Thus, if we write $\widehat{q} = \beta^{-1}(q)$
to denote a preimage of any $q\in Q$ under
$\beta$, and $\widehat{k} = \alpha(k)$ be the image
of any $k\in K$ under $\alpha$, then 
our definition says that 
$\widehat{\theta(q)(k)} = \widehat{q}\;\widehat{k}\;\widehat{q}^{-1}$.

It is not immediately clear whether equation (\ref{eqn})
defines a function, because there is some freedom
in choosing a preimage $\beta^{-1}(q)$. However,
one can prove that different choices for
$\beta^{-1}(q)$ lead to the same value for
$\theta(q)(k)$ when $K$ is abelian. (This is
explained in a remark below.)

\begin{thm}
If $\displaystyle 
0\rightarrow K \stackrel{\alpha}{\rightarrow} G
\stackrel{\beta}{\rightarrow} Q\rightarrow 1$
and 
$\displaystyle 
0\rightarrow K \stackrel{\alpha'}{\rightarrow} G'
\stackrel{\beta'}{\rightarrow} Q\rightarrow 1$
are equivalent, then they realize the
same homomorphism
$\theta=\theta':Q\rightarrow {\rm Aut}(K)$.
\end{thm}

\begin{proof}
%Assume that the first extension realizes
%$\theta$ and the second relaizes $\theta'$.
Recall that the two extensions are equivalent
if there is an isomorphism $\gamma:G\rightarrow G'$
such that $\gamma\circ\alpha = \alpha'$ and
$\beta'\circ\gamma = \beta$.
We must use these facts to show that 
for any $q\in Q$ and $k\in K$ we have
$\theta(q)(k) = \theta'(q)(k)$.
As above, let $\widehat{k} = \alpha(k)$
and let $\widehat{q}$ be some element of 
$\beta^{-1}(q)$. Let 
$\widehat{\widehat{k}} = \alpha'(k)$
and let $\widehat{\!\widehat{q}}$ be any element
of $(\beta')^{-1}(q)$. We must show that 
\[
\alpha^{-1}\left(\widehat{q}\;\widehat{k}\;\widehat{q}^{-1}\right) 
\stackrel{?}{=}
(\alpha')^{-1}\left(\widehat{\!\widehat{q}}\;\widehat{\widehat{k}}
\;\widehat{\!\widehat{q}}^{-1}\right).
\]
Let's rewrite this in
an equivalent but simpler form.
Since $\gamma\circ\alpha = \alpha'$ is 1-1, 
applying this function to both sides
produces an equivalent line:
\[
\gamma\alpha\alpha^{-1}\left(\widehat{q}\;\widehat{k}\;\widehat{q}^{-1}\right)
\stackrel{?}{=}
(\alpha')(\alpha')^{-1}\left(\widehat{\!\widehat{q}}\;\widehat{\widehat{k}}
\;\widehat{\!\widehat{q}}^{-1}\right),
\]
or
\[
\gamma\left(\widehat{q}\;\widehat{k}\;\widehat{q}^{-1}\right) 
\stackrel{?}{=}
\widehat{\!\widehat{q}}\;\widehat{\widehat{k}}\;
\widehat{\!\widehat{q}}^{-1}.
\]
But since 
$\gamma(\widehat{k}) = \widehat{\widehat{k}}$,
this reduces to 
\[
\gamma\left(\widehat{q}\;\widehat{k}\;\widehat{q}^{-1}\right) =
\gamma(\widehat{q})\;\gamma(\widehat{k})\;
\gamma(\widehat{q}^{-1}) =
\gamma(\widehat{q})\;\widehat{\widehat{k}}\;
\gamma(\widehat{q}^{-1}) \stackrel{?}{=}
\widehat{\!\widehat{q}}\;\widehat{\widehat{k}}\;
\widehat{\!\widehat{q}}^{-1}.
\]
The equation $\beta'\circ\gamma = \beta$ implies that 
$\beta'(\gamma(\widehat{q})) = \beta'(\widehat{\!\widehat{q}})$,
so there is some $j\in K$ such that 
$\gamma(\widehat{q})\alpha'(j)
= \widehat{\!\widehat{q}}$.
Substituting this on the righthand side of the
previous line we get 
an equivalent line
\[
\gamma(\widehat{q})\;\widehat{\widehat{k}}\;
\gamma(\widehat{q}^{-1}) \stackrel{?}{=}
\gamma(\widehat{q})\;\alpha'(j)\;
\widehat{\widehat{k}}\;
(\alpha'(j))^{-1}\;
\gamma(\widehat{q}^{-1}),
\]
so (after canceling identical factors
on the right and left)
the line we have to verify reduces to 
\[
\widehat{\widehat{k}} \stackrel{?}{=}
\alpha'(j)\;\widehat{\widehat{k}}\;
(\alpha'(j))^{-1}.
\]
Since $\widehat{\widehat{k}}=\alpha'(k)$ we
can rewrite this as
\[
\alpha'(k)\alpha'(j) \stackrel{?}{=}
\alpha'(j)\alpha'(k),
\]
or
$\alpha'(kj) \stackrel{?}{=} \alpha'(jk)$.
Since $K$ is abelian, $kj=jk$, so we are done.
\end{proof}

\noindent
{\bf Remark.} A special case of this theorem 
is when $G = G', \alpha = \alpha', \beta = \beta'$ 
and $\gamma = \id_G$. In this case, if one does NOT
assume that $\beta^{-1}(q) = (\beta')^{-1}(q)$,
then one has a proof that different choices for
$\beta^{-1}(q)$ lead to the same value
of $\theta(q)(k)$; that is, one has proof that
$\theta$ is well defined.

This theorem shows that there is a 
well defined way 
of assigning to an equivalence
class of extensions of $K$ by $Q$
a homomorphism $\theta:Q\rightarrow {\rm Aut}(K)$.
The purpose of these notes
is to show that this assignment is
onto. For each $\theta:Q\rightarrow {\rm Aut}(K)$
there is an extension $G = K\rtimes_{\theta} Q$
and homomorphisms $\alpha$ and $\beta$ such that
the extension
\[
0\rightarrow K \stackrel{\alpha}{\rightarrow} K\rtimes_{\theta} Q
\stackrel{\beta}{\rightarrow} Q\rightarrow 1
\]
realizes $\theta$.

\begin{df} 
Assume that $K = \lb K; +, -, 0_K\rb$ and 
$Q = \lb Q; \cdot, {}^{-1},1_Q\rb$ are groups, and 
that $\theta:Q\rightarrow {\rm Aut}(K)$
is a homomorphism. Write
${}^qk$ as a shorthand for $\theta(q)(k)$. The {\it semidirect
product of $K$ by $Q$} is the algebra
$K\rtimes_{\theta} Q=\lb K\times Q; *, {}^{-1}, 1\rb$
where
\begin{enumerate}
\item[(1)] $(k,q)*(j,p) = (k + {}^qj,q\cdot p)$;
\item[(2)] $(k,q)^{-1} = \left({}^{q^{-1}}\left(k^{-1}\right),q^{-1}\right)$;
and
\item[(3)] $1_{K\rtimes Q} = (0_K,1_Q)$.
\end{enumerate}
\end{df}

%\vfill
%\eject

\begin{thm} The following hold.
\begin{enumerate}
\item[(1)] $K\rtimes_{\theta} Q$ is a group.
\item[(2)] The functions $\alpha:K\rightarrow K\rtimes_{\theta} Q:
k\mapsto (k,1)$ and $\beta:K\rtimes_{\theta} Q\rightarrow Q:(k,q)\mapsto
q$
are homomorphisms.
\item[(3)] The sequence 
$0\rightarrow K \stackrel{\alpha}{\rightarrow} K\rtimes_{\theta} Q
\stackrel{\beta}{\rightarrow} Q\rightarrow 1$ is exact,
and realizes $\theta$.
%\item[(4)] The extension
%$0\rightarrow K \stackrel{\alpha}{\rightarrow} K\rtimes_{\theta}
%Q \stackrel{\beta}{\rightarrow} Q\rightarrow 1$ realizes $\theta$.
\end{enumerate}
\end{thm}

\begin{proof}
See the proof of Theorem 7.22 from the book.
\end{proof}

\begin{thm} (Characterization of semidirect products)
The following are equivalent.
\begin{enumerate}
\item[(1)] $G\cong K\rtimes_{\theta} Q$ for some $\theta$.
\item[(2)] $G$ has subgroups $K'$ and $Q'$ such that
\begin{enumerate}
\item[(i)] $K'\cong K$ and $Q'\cong Q$;
\item[(ii)] $K'\lhd G$; 
\item[(iii)] $K'\vee Q' = G$; and
\item[(iv)] $K'\cap Q' = \{1\}$.
\end{enumerate}
\end{enumerate}
\end{thm}

\begin{proof}
This is essentially Theorem 7.23 of the book.
\end{proof}

\noindent
{\bf Examples.}
\begin{enumerate}
\item[(a)] A direct product is a semidirect product:
$K\times Q = K\rtimes_{\theta} Q$ for $\theta:Q\rightarrow 
{\rm Aut}(K): q\mapsto 1$, the constant homomorphism.
\item[(b)] $D_n \cong Z_n\rtimes_{\theta} {\BZ}_2$.
\item[(c)] $S_n \cong A_n\rtimes_{\theta} {\BZ}_2$.
\item[(d)] $S_4\cong V\rtimes S_3$.
\item[(e)] $GL(n,{\BF})\cong SL(n,{\BF})\rtimes F^*$, where
$F^*$ is the group of invertible diagonal matrices
where $a_{ii} = 1$ for all $i>1$. (The entry
$a_{11}$ can be any nonzero element of ${\BF}$.)
\item[(f)] The unique nonabelian group of order 21,
$\lb a, b\;|\;a^7 = 1, b^3 = 1, ba = a^2b\rb$, 
is a semidirect product. It is isomorphic to 
${\BZ}_7\rtimes_{\theta} {\BZ}_3$ for
$\theta:{\BZ}_3\rightarrow {\rm Aut}({\BZ}_7): 
\theta(1)(x)=2x$.
(Can you see why?)
\end{enumerate}
\end{document}

