Presentations representing different groups.

Question

Using GAP, I knew that groups $G=\langle x,y;x^4,x^2y^2,xyxy^{-1}\rangle$ and $H=\langle x,y;x^4,y^4,xyxy^{-1}\rangle$ are different. But I want to prove it.

I tried to do something using Tietze transformations but could not get success. Please suggest me a proof.

Answer 1

Hint: If the groups were isomorphic, then their abelianisations would be isomorphic as well. That is, $G\cong H$ implies $G/[G,G]\cong H/[H,H]$. Check that $H/[H,H]$ has an element of order $4$, while $G/[G,G]$ does not. (Just add the relation $xy = yx$ to each presentation to compute the abelianisation.)

Answer 2

Well $H$ is visibly a semidirect product of two cyclic of order $4$, so it has order $16$.

But $G$ has a normal subgroup $\langle x \rangle$ of order at most $4$, and $y^2 \in \langle x \rangle$, so $|G| \le 8$. (In fact $|G|=8$.)