TITLE: $G$ abelian $\{ a^ib^j : a^m = b^n = 1, a^2=b^3 \}$, prove that $G=\langle c:c^k=1 \rangle$
QUESTION [6 upvotes]: Let G be an abelian group $\{ a^i b^j : a^m = b^n = 1, a^2=b^3 \}$. Prove that $G=\langle c:c^k=1 \rangle$ with $k=\gcd(3m,2n)$. Hint: consider the element $ab^{-1}$.

This should be easy but I'm having a hard time with it. I think even with the hint that $ab^{-1}$ is supposed to be the generator $c$. 
So that point it to prove that all the powers of $c$ represent every element in the group. My first idea was to notice that $a^{2m}b^{3n}=(ab)^h=1$ where $h=lcm(2m,3n)$. So $(ab)^h$ could be a generator, but I am not sure if all the elements of $G$ are represented, and it seems it does not use the hint. 
I could use some help on this. Thanks.
OK so with the (other) hint, $(ab^{-1})^2=b$ and $(ab^{-1})^3=a$, so $(ab^{-1})^{2n}=b^n=1$ and $(ab^{-1})^{3m}=a^m=1$, that is how we get the $3m$ and $2n$. $k=\gcd(3m, 2n)$, and I find then there exists 2 integers, $h$ and $g$ such that $kh=3m \implies k=\frac{3m}{h}$ and $kg=2n \implies k=\frac{2n}{g}$. However still stuck here.

REPLY [2 votes]: Set $c = ab^{-1}.$  Then obviously $c \in G,$ so $\langle c\rangle \subseteq G$ but also since $c^2 = b$ and $c^3 = a,$ we have $G = \langle a,b\rangle \subseteq \langle c \rangle$ so $\langle c\rangle = G.$
The group presentation is then
$$\begin{align*}G &= \langle a, b  \mid a^m = b^n = 1, a^2 = b^3\rangle \\ &= \langle c \mid c^{3m} = c^{2n} = 1, c^6=c^6\rangle \\ &= \langle c \mid c^{3m}=c^{2n}=1\rangle \\ &= \langle c \mid c^{\gcd(3m,2n)} = 1\rangle\end{align*}$$