Is $\langle a,b | a^2b^2=1 \rangle$ a semidirect product of $\mathbb{Z}^2$ and $\mathbb{Z}_2$?

Question

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.

Answer 1

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 $Z_2$.

Answer 2

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 $BS(1,-1)$.

Answer 3

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...

Answer 4

No, because your group doesn't have any element of order $2$, so can't contain $\mathbb{Z}_2$.

To see this : using the fact that $b^2=a^{-2}$, one sees that every element of your group can be written uniquely as : $$a^{n_1} b^{n_2} a^{n_3} \dots a^{n_{r-1}} b^{n_r},$$ with $r$ an even positive integer, $n_i \in \mathbb{Z}_{\neq 0}$ if $i$ is odd and $n_i \in \{-1,1\}$ if $i$ is even, and eventually $n_1=0$ or $n_r=0$.

Now suppose that an element $x = a^{n_1} b^{n_1} a^{n_2} \dots a^{n_r} b^{n_r}$ of your group is of order $2$. Then the string : $$a^{n_1} b^{n_2} a^{n_3} \dots a^{n_{r-1}} b^{n_r} \cdot a^{n_1} b^{n_2} a^{n_3} \dots a^{n_{r-1}} b^{n_r}$$ cancels to $1$. This implies $$ n_r=0, $$ $$ n_{r-1}=n_1, $$ $$ n_{r-2}=n_2, $$ $$ ... $$ So in particular : $n_{r/2} = -n_{r/2}$ wich implies $r=0$.