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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.

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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 $y \in \mathbb Z_2$? – Rudy the Reindeer 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 book "Combinatorial Group Theory". – user1729 Sep 10 '12 at 10:39
up vote 4 down vote accepted

Changing $b$ to $b^{-1}$ you can rewrite the presentation as $$G=<a, b| a^2=b^2>$$ 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 $G=Z*_Z Z$ 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 $<A,B|A^2=B^2=1>$ which is the infinite dihedral group $Q=Z_2*Z_2$. The group $Q$ is has an infinite cyclic subgroup $K$ generated by $AB$, with quotient $Z_2$; it is a semidirect product of $K=Z$ by $Z_2$.

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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...

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