TITLE: Is $\langle a,b \mid a^2b^2=1 \rangle$ a semidirect product of $\mathbb{Z}^2$ and $\mathbb{Z}_2$?
QUESTION [10 upvotes]: All is in the title: Is $\langle a,b \mid a^2b^2=1 \rangle$ 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.

REPLY [5 votes]: Changing $b$ to $b^{-1}$ you can rewrite the presentation as $$G=\langle a, b\mid a^2=b^2\rangle.$$ 
The group  is not a semidirect product since the group $G$ does not have non-trivial elements of finite order. One way to see that is to realize that $G=\mathbb{Z}*_{\mathbb{Z}} \mathbb{Z}$ is a free product with amalgamation of two infinite cyclic groups generated by $a, b$ amalgamated along their subgroups $a^2, b^2$. The elements of finite order of such a free product with amalgamation must be conjugate to one of the factors, so there is no torsion.
The element $a^2$ ( or $b^2$) is central and generates an infinite cyclic subgroup $C$. Then $Q=G/C$ has the presentation $\langle A,B\mid A^2=B^2=1\rangle$ which is the infinite dihedral group $Q=\mathbb{Z}_2*\mathbb{Z}_2$. The group $Q$ is has an infinite cyclic subgroup $K$ generated by $AB$, with quotient $\mathbb{Z}_2$; it is a semidirect product of $K=\mathbb{Z}$ by $\mathbb{Z}_2$.