Free group with relations I understand that $\langle a\vert a^2\rangle$ is isomorphic to $\mathbb{Z}_2$.
My question is $\langle a,b,c\vert a^2b^2c^2\rangle$ isomorphic to $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$?
Thanks.
 A: As far as I know, regarding to $\langle a|a^2=1\rangle\cong\mathbb{Z_2}$, $$\langle a,b,c|a^2=b^2=c^2=1, [a,b]=[a,c]=[b,c]=1\rangle\cong\mathbb{Z_2}\times\mathbb{Z_2}\times\mathbb{Z_2}$$
A: Nope! We can collapse $\langle a, b, c|a^2b^2c^2\rangle \to \langle a, b, c|a^2b^2c^2, c=ab=ba=e\rangle=\langle a, b|ab=ba=e\rangle=\mathbb{Z}$. Clearly this is surjective. Thus we have a surjection onto $\mathbb{Z}$, so the group is not even finite! Note since we factored through the abelization, even commutativity is not a problem, this is just a problem of insufficient relations.
Edit:
As for the collapsing part, it is acutally pretty straightforward. If you have a free group on three letters, $F_3$, and denote the normal subgroup generated by $\{a^2b^2c^2\}=I$ and $\{a^2b^2c^2, [a,b], c, ab\}=J$, then the group you are talking about is $F_3/I$ by definition. Now I am talking about $F_3/J$. Now we have $I\subseteq J$, which means that $J/I$ is normal subgroup of $F_3/J$, so we have a map $F_3/J\to (F_3/I)/(J/I)=F_3/J$ by Noether's Third theorem. This actually is a lot more general! If we talking about some $n$-letters, for some cardinal, with relations $R\subseteq Q$, then the exact same aurguments creates a map $\langle n| R\rangle \to \langle n| Q\rangle $.
