$G$ abelian $\{ a^ib^j : a^m = b^n = 1, a^2=b^3 \}$, prove that $G=\langle c:c^k=1 \rangle$ 
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.
 A: We can find out how to express our finitely generated group $G$ as a direct product of cyclic groups by finding the Smith normal form of
$$\begin{bmatrix} m & 0 & 2 \\ 0 & n & -3 \end{bmatrix}.$$
Let's perform the following row and column operations: first adding $2\cdot\text{Row }1$ to $\text{Row }2$, we have
$$\begin{bmatrix} m & 0 & 2 \\ 2m & n & 1 \end{bmatrix}.$$
Switching the rows, we have
$$\begin{bmatrix} 2m & n & 1 \\ m & 0 & 2 \end{bmatrix}.$$
Switching the first and last columns, we have
$$\begin{bmatrix} 1 & n & 2m \\ 2 & 0 & m \end{bmatrix}.$$
Adding $-2\cdot\text{Row }1$ to $\text{Row }2$, we have
$$\begin{bmatrix} 1 & n & 2m \\ 0 & -2n & -3m \end{bmatrix}.$$
Adding $-n\cdot\text{Column }1$ to $\text{Column }2$, we have
$$\begin{bmatrix} 1 & 0 & 2m \\ 0 & -2n & -3m \end{bmatrix}.$$
Adding $-2m\cdot\text{Column }1$ to $\text{Column }3$, we have
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & -2n & -3m \end{bmatrix}.$$
Finally, multiplying $\text{Row }2$ by $-1$, we have
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2n & 3m \end{bmatrix}.$$
This last matrix will have the same Smith normal form as our original matrix. Let $k=\text{GCD}(2n,3m)$. Then the Smith normal form will be
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & k & 0 \end{bmatrix}.$$
It follows that $G$ is a cyclic group of order $k$.
A: 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
$$\begin{align*}G &= \langle a, b  \mid a^m = b^n = 1, a^2 = b^3\rangle \\ &= \langle c \mid c^{3m} = c^{2n} = 1, c^6=c^6\rangle \\ &= \langle c \mid c^{3m}=c^{2n}=1\rangle \\ &= \langle c \mid c^{\gcd(3m,2n)} = 1\rangle\end{align*}$$
