What is the abelianization of $\langle x,y,z\mid x^2=y^2z^2\rangle?$

Solution 1:

You can rewrite your relator such that it has $0$ exponent sum in two of the generators, as the map $x\mapsto xyz, y\mapsto y, z\mapsto z$ is a Nielsen transformation: $$ \begin{align*} \langle x, y, z\mid x^{2}=y^2z^2\rangle &\cong\langle x, y, z\mid x^{2}z^{-2}y^{-2}\rangle\\ &\cong\langle x, y, z\mid (xyz)^{2}z^{-2}y^{-2}\rangle \end{align*} $$ Under the abelinisation map we then get the group: $$ \begin{align*} \langle x, y, z\mid (xyz)^{2}z^{-2}y^{-2}\rangle^{ab}&=\langle x, y, z\mid x^2\rangle^{ab}\\ &\cong \mathbb{Z}^2\times(\mathbb{Z}/2\mathbb{Z}) \end{align*} $$

This is a specific case of a more general phenomenon, where one can adapt the Euclidean algorithm to rewrite using automorphisms a word $W\in F(a, b, \ldots)$ such that it has zero exponent sum in all but one of the relators. For example, writing $\sigma_x$ for the exponent sum of the relator word in the letter $x$: $$ \begin{align*} &\langle a, b\mid a^6b^8\rangle&&\sigma_a=6, \sigma_b=8\\ &\cong\langle a, b\mid (ab^{-1})^6b^8\rangle&&\text{by applying}~a\mapsto ab^{-1}, b\mapsto b\\ &=\langle a, b\mid (ab^{-1})^5ab^7\rangle&&\sigma_a=6, \sigma_b=2\\ &\cong\langle a, b\mid (a(ba^{-3})^{-1})^5a(ba^{-3})^7\rangle&&\text{by applying}~a\mapsto a, b\mapsto ba^{-3}\\ &\cong\langle a, b\mid (a^4b^{-1})^5a(ba^{-3})^7\rangle&&\sigma_a=0, \sigma_b=2 \end{align*} $$ You can think of this as a "non-commutative Smith normal form", but it is more useful in this context than the Smith normal form as it gives you more information than just the abelianisation. For example, it is used in the HNN-extension version of the Magnus hierarchy ($a$ is the stable letter, and the associated subgroups are free by the Freiheitssatz; see J. McCool and P. Schupp, On one relator groups and HNN extensions, Journal of the Australian Mathematical Society, Volume 16, Issue 2, September 1973 , pp. 249-256 doi).

Solution 2:

I would think of it as starting with $\Bbb{Z}^3=\langle x,y,z\rangle $ and then quotienting out the cyclic subgroup $\langle x^{-2}y^2z^2\rangle$. You can see easily that $x^{-1}yz$, $y$, and $z$ are an alternate set of generators for $G$, so you get $\Bbb{Z}^2\times \Bbb{Z}/2\Bbb{Z}$ generated by $y, z,$ and $x^{-1}yz$ when you take the quotient.

There is no need to know anything about non-abelian groups, since you can abelianize first and then mod out by the relation. (Abelianization is just modding out by some relations, and it doesn't matter which relations you mod out first--you get to the same place in the end.)