Is $\langle a,b \mid a^2b^2=1 \rangle$ a semidirect product of $\mathbb{Z}^2$ and $\mathbb{Z}_2$? 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.
 A: 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 $\mathbb{Z}_2$.
A: If you want a semidirect product of $\mathbb{Z}$ and $\mathbb{Z}_2$ an extra relation is required: you need $a^2=1=b^2$. Then the presentation gives the infinite dihedral group (i.e. the semidirect product of the question). 
To see this put $t=ab$. Then if we call your group $G$, we have $G\cong\langle a,t\rangle$ and the defining relation becomes $ata^{-1}=t^{-1}$. It now follows that $\langle t\rangle$ is normal in $G$ with $\langle a\rangle$ acting on $\langle t\rangle$ by conjugation with kernel $\langle a^2\rangle$. With the extra relation above, this kernel is trivial. Thus $G\cong\langle t\rangle\rtimes\langle a\rangle\cong\mathbb{Z}\rtimes\mathbb{Z}_2$.
Without the extra relation the group is isomorphic to $\langle a,t\ |\ ata^{-1}=t^{-1}\rangle$, which is also known as the Baumslag-Solitar group $BS(1,-1)$.
A: Yes...if what you actually meant was $\,\langle\,a\,,\,b\;|\;a^2=b^2=1\,\rangle\,$. This is $\, C_2*C_2=\,$ the free product of two groups of order two, also known as the infinite dihedral group. I think yours is missing the relator $\,b^2=1\,$
Every presentation of such a group gives some different interesting insights in its structure...
