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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.

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    I looked up the [definition on Wikipedia](http://en.wikipedia.org/wiki/Semidirect_product) 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 $y \in \mathbb Z_2$?2012-09-05
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    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 $\langle a,b |a^2b^2=1 \rangle$.2012-09-05
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    We can notice that $BS(1,-1) = \mathbb{Z} \rtimes \mathbb{Z}$.2012-09-05
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    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 book "Combinatorial Group Theory".2012-09-10
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    @user1729: I can't find the proof in that book. Would you be more specific, please?2018-05-15
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    @Shaun search for a paper "On one relator groups and HNN extensions", by McCool and Schupp. The proof in there is much nearer (avoiding staggered presentations). The proof in Magnus, karrass and solitar is in the section on one-relator groups.2018-05-15
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    *neater, not nearer2018-05-15

3 Answers 3

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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 $\mathbb{Z}_2$.

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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 $BS(1,-1)$.

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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...