TITLE: What is the abelianization of $\langle x,y,z\mid x^2=y^2z^2\rangle?$
QUESTION [11 upvotes]: Let $G=\langle x,y,z\mid x^2=y^2z^2\rangle$.
What is the abelianization of this group?
(Also, is there a general method to calculate such abelianizations?)
Update: I know how to get a presentation of the abelianization by adding relations like $xy=yx$ and so on. However is it possible to express it as a direct sum of cyclic groups as per the fundamental theorem for finitely generated abelian groups?
Thanks.

REPLY [3 votes]: If we have
$$\langle x,y,z\ |\ x^{-2}y^2z^2=1\rangle^{\rm ab}=\langle x,y,z\ |\ x^{-2}y^2z^2=[x,y]=[x,z]=[y,z]=1\rangle$$
with one of the Tietze moves
$$\langle x,y,z,t\ |\ t=x^{-1}yz, t^2=[x,y]=[x,z]=[y,z]=1\rangle$$
With another, now we arrange
$$\langle x,y,z,t\ |\ x=yzt^{-1}, t^2=[x,y]=[x,z]=[y,z]=1\rangle$$
and finally get
$$\langle y,z,t\ |\ t^2=[y,z]=1\rangle$$
which clearly is $\Bbb Z+\Bbb Z+\Bbb Z_2$.