For what $m,n$ is $D_{mn}$ isomorphic to $D_m \times Z_n$? For what $m,n$ is $D_{mn}$ isomorphic to $D_m \times Z_n$?
Should I try to define an isomorphism between them? Where to start?
 A: Recall that $D_n\cong \Bbb Z_n\rtimes\Bbb Z_2$.  So we want to know when $\Bbb Z_m\times(\Bbb Z_n\rtimes\Bbb Z_2)\cong \Bbb Z_{mn}\rtimes\Bbb Z_2$.  This will be true when $\Bbb Z_{mn}\cong\Bbb Z_m\times\Bbb Z_n$, or , when $(m,n)=1$.
A: $\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Let $D_{k} = \Span{ a, b : a^{k} = 1, b^{2} = 1, b^{-1} a b = a^{-1} }$. When is it that $D_{k}$ has a non-trivial (cyclic) direct factor?
If $k = 1$, then $D_{1} \cong Z_{2}$. If $k = 2$, then $D_{ 2} \cong Z_{2} \times Z_{2} \cong D_{1} \times Z_{2}$.
If $k \ge 3$, it is easy to see that no element $c$ outside $\Span{a}$ is in the centre, as $c^{-1} a c = a^{-1} \ne a$. Now if an element $a^{t} \ne 1$ is in the centre, we must have $a^{-t} = b^{-1} a^{t} b = a^{t}$, hence $a^{2 t} = 1$, that is, $a^{t}$ is an involution, which can only happen when $k$ is even, and then $a^{t} = a^{k/2}$. And in this case $a^{k/2}$ is a direct factor in $\Span{a}$, and thus in $D_{k}$, if and only if $k/2$ is odd,
The moral of the story is that this happens iff $n = 2$ and $m$ is odd.
