If $G$ is a group and $N$ is a normal subgroup and if $G=\langle x,y \rangle$ then $G/N = \langle xN,yN \rangle$.
I am not very comfy with quotient groups so this problem has had me stumped for some days.
I know elements of $G$ can be written of the form $x^my^n$ for some $m,n \in \Bbb{Z}$. How do I prove elements of the quotient group can be generated by $xN,yN$?
BTW sorry if this is a duplicate, I couldn't find one.
The elements of $G$ might not necessary be written in the form $x^my^n$. In general, is $S\subseteq G$ is a set and $G=\langle S\rangle$ then a general element of $G$ has the form:
$s_1^{n_1}s_2^{n_2}...s_k^{n_k}$ where $s_1,...,s_k\in S, n_1,...,n_k\in\mathbb{Z}, k\geq 0$
So in your case a general element has the form $s_1^{n_1}s_2^{n_2}...s_k^{n_k}$ where $s_1,...,s_k\in\{x,y\}$ and the powers are integers. Since $G$ is not necessary abelian, you in general can't change the order and turn this element into $x^my^n$.
Now, the operation of $G/N$ is simply $(g_1N)(g_2N)=g_1g_2N$. A general element in the group has the form $gN$ for some $g\in G$. As noted above, we can write $g=s_1^{n_1}s_2^{n_2}...s_k^{n_k}$ like before. Then:
$gN=s_1^{n_1}s_2^{n_2}...s_k^{n_k}N=(s_1N)^{n_1}(s_2N)^{n_2}...(s_kN)^{n_k}$
This is a product of the elements $xN, yN$ and their inverses, so it indeed belongs to $\langle xN, yN\rangle$.