Describing groups with given presentation? $\langle x,y\ |\ xy=yx,x^5=y^3\rangle$ and $\langle x,y\ |\ xy=yx,x^4=y^2\rangle$. I'm trying to describe the groups with presentations $\langle x,y\ |\ xy=yx,x^5=y^3\rangle$ and $\langle x,y\ |\ xy=yx,x^4=y^2\rangle$. I have some problems getting a good picture of what they look like... 
For the first one, $xy=yx$ allows us to say that any element in the group can be written as $x^iy^j$ for some $i,j$. More precisely, using $x^5=y^3$, we can always reduce $j$ to $0,1$ or $2$, whereas $i$ could have any value. So I was thinking about $\mathbb{Z} \times C_3$. But the other way around (reducing $i$ this time) it's also $C_5 \times \mathbb{Z}$. But then I should be able to prove that both descriptions are equivalent, which I can't do...
For the second, I proceed similarly and get to the same kind of problem...
Could you tell me where I'm going wrong?
Thank you very much in advance!
 A: The group $\langle x,y | xy=yx, x^5=y^3 \rangle$ is the quotient of $\langle x,y | xy=yx \rangle = \mathbb{Z} \oplus \mathbb{Z}$ (with $x=(1,0)$ and $y=(0,1)$) by the (normal) subgroup generated by $5x-3y=(5,-3)$, and similarly $\langle x,y | xy=yx , x^4=y^2 \rangle$ is the quotient of $\mathbb{Z} \oplus \mathbb{Z}$ by the (normal) subgroup generated by $(4,-2)$. Now in general we have the result for $0 \neq (n,m) \in \mathbb{Z} \oplus \mathbb{Z}$, that
$$(\mathbb{Z} \oplus \mathbb{Z})/\langle (n,m) \rangle \cong \mathbb{Z} \oplus \mathbb{Z}/\langle d \rangle,$$
where $d := \mathrm{gcd}(n,m)$. I am sure that this has been explained a couple of times on math.SE, but here you can also see it "directly":
Write the first group additively as $\langle x,y : 5x=3y \rangle$ (the commutativity being implicit; i.e. we take a presentation as an abelian group). We may rewrite the relation as $2x=3(y-x)$, i.e. $2x=3y'$ after replacing $y$ by $y'+x$. The new relation can be written as $2(x-y')=y'$, i.e. $2x'=y'$ after replacing $x$ by $y'+x'$. But now $y'$ is superfluous, and we see that the group is freely generated by $x'=2x-y$.
Does this remind you of the Euclidean algorithm? That's exactly what happens in the Smith normal form for $1 \times 2$-matrices, and the Smith normal form lets you decompose finitely generated abelian groups into direct sums of cyclic groups.
