Take the tour ×  All is in the title: Is $\langle a,b | a^2b^2=1
\rangle$ a semidirect product of $\mathbb{Z}^2$ and
$\mathbb{Z}_2$? I think it is the case, but I don't know how to
prove it.  share|improve this question I looked up the
definition on Wikipedia and it says that $G$ is the semidirect
product of normal subgroups $N,H \subset G$ if $G = NH$ and $N
\cap H = \{0\}$. How would you define multiplication $x \cdot y$
where $x \in \mathbb Z^2$ and $y \in \mathbb Z_2$? –  Matt N.
Sep 5 '12 at 8:57 It is the main problem: if we know how
$\mathbb{Z}_2$ acts on $\mathbb{Z}^2$, we could compare the
presentation of the semidirect product with $\langle a,b
|a^2b^2=1 \rangle$. –  Seirios Sep 5 '12 at 9:02 We can notice
that $BS(1,-1) = \mathbb{Z} \rtimes \mathbb{Z}$. –  Seirios Sep
5 '12 at 15:10 Your group has a single defining relator,
$a^2b^2$. A group with a single defining relator has torsion if
and only if the relator is a proper power. One can find a proof
of this fact in Magnus, Karrass and Solitar's boo