Let $a, b\in\mathbb Z$, and assume that $\gcd(a,b)=1$.
a) Show that there exist $x_0$, $y_0\in\mathbb Z$ such that $ax_0 + by_0 = 1$.
b) Suppose $m\in\mathbb Z$ and $a\mid bm$. Show that $a\mid m$. (Hint: multiply the equation in part (a) by the number $m$.)
c) Unrelated to part (b): suppose $(G, \cdot)$ is an abelian group, and $g, h\in G$ have orders $a$ and $b$ respectively. What can you say about $(gh)^{ax_0}$ ? Where did you use commutativity?
I solved parts a) and b), and here is my attempt for part c):
$g^a = 1$ and $h^b = 1$ since $g, h\in G$ have orders $a$ and $b$ respectively.
$$\begin{align}(gh)^{ax_0} &= (gh)(gh)\cdots(gh) \\&= (gg\cdots g).(hh\cdots h) \\&= g^{ax_0}.h^{ax_0} \\&= (g^a)^{x_0}\cdot h^{1-by_0} \\&=1\cdot h\cdot h^{-by_0} \\&= 1\cdot h\cdot (h^b)^{-y_0} \\&= 1\cdot h\cdot 1 \\&= h\end{align}$$ (since G is commutative)
Is this correct? and if so, what does this result tell me, i.e. what can I say about $(gh)^{ax_0}=h$?
Yes, you're correct, and you used the commutativity at the equality indicated
$$\begin{align}(gh)^{ax_0} &= (gh)(gh)\cdots(gh) \\&\color{red}{\stackrel{\text{here}}{=}}(gg\cdots g).(hh\cdots h) \\&= g^{ax_0}.h^{ax_0} \\&= (g^a)^{x_0}\cdot h^{1-by_0} \\&=1\cdot h\cdot h^{-by_0} \\&= 1\cdot h\cdot (h^b)^{-y_0} \\&= 1\cdot h\cdot 1 \\&= h.\end{align}$$
One can say that one can cancel the $h$ on the right to get $$(gh)^{ax_0-1}\cdot g=e_G.$$ However, your question is somewhat open-ended, so it's difficult to answer with sufficient accuracy.