I'm trying to help a friend through his Maths degree and he gave my a question that I just can't remember how to do. Show $$\langle x,y | y^5=x^2=[x,y]=1\rangle \ \text{and}\ \langle u, w | u^{-1} wu=w^2, w^{-1}uw=u^2\rangle$$
are isomorphic and justify. I recall that there's some work to be done with generators etc but I can't find questions similar enough that I would feel confident helping him without mistakes.
I don't think they are isomorphic... Call the first group $G$ and the second group $H$.
The first group is abelian, since $x$ and $y$ commute; it has an element of order $2$ and an element of order $5$, hence it has an element of order $10$, namely $xy$. The group is, in fact, cyclic of order $10$: if $G=C_{10}$ generated by $z$, then $z^5$ and $z^2$ satisfy the relations given for $x$ and $y$, so we get a map $G\to C_{10}$ given by $x\mapsto z^5$ and $y\mapsto z^2$. The map is surjective, so $G$ has order at least $10$. Conversely, we can map $C_{10}$ to $G$ by sending $z$ to $xy$, which has order $10$ since $[x,y]=1$; the map is surjective, since $(xy)^2 = y^2$ and $\langle y^2\rangle =\langle y\rangle$ (as the order of $y$ divides $5$), so $\langle xy\rangle$ contains $y$ and hence contains $x$. This proves $G$ has order at most $10$, and so $G\cong C_{10}$.
On the other hand, if $H$ were abelian, then $u^{-1}wu=w^2$ would imply $w=w^2$, so $w=e$, and similarly $w^{-1}uw=u^2$ would imply $u=e$, and so we would immediately conclude that if $H$ is abelian, then it is trivial. Thus, either $H$ is not abelian (and so cannot be isomorphic to $G$), or else $H$ is abelian and trivial (and so cannot be isomorphic to $G$).
So, what is $H$ isomorphic to? Note that $[u,w]=u^{-1}w^{-1}uw=u$ and $[w,u]=w^{-1}u^{-1}wu = w$. Since $[u,w]=[w,u]^{-1}$, this means $w=u^{-1}$. But then $u^2 = w^{-1}uw = uuu^{-1}$, so $u^2=u$, hence $u=e$ and similarly $w=e$, so $H$ is in fact trivial.