I need to show that the group $G$ generated by $(x,y)$ such that $xyx^{-1}y^{-1}=1$ is isomorphic to the group $H$ of ordered pairs of integers under addition. My first thought is to use the first isomorphism theorem. To do this, I need to find a surjective homomorphism from some group $G'$ onto $H$ such that $G$=$\frac{G'}{N}$, where $N$ is the kernel.
I don't know how to find such a homomorphism, or if there is perhaps a better way of doing this. Any help would be greatly appreciated.
Suppose that you have an element of $G$. Then it will have the form $x^{n_1}y^{k_1}x^{n_2}y^{k_2}\cdots x^{k_i}y^{k_i}$ for some $i$ by the definition of $x$ and $y$ generating $G$. Since, the elements commute, we actually get that this is equal to $x^{n_1+\cdots+n_i}y^{k_1+\cdots+k_i}$. This means that all elements are of the form $x^ny^k$ for some $n,k \in \Bbb{Z}$. In fact, each element can be written uniquely this way.
Now, to construct an isomorphism from $G$ to $\Bbb{Z} \times \Bbb{Z}$, notice that all elements of $\Bbb{Z} \times \Bbb{Z}$ are of the form $(n,k)$ for some $n,k \in \Bbb{Z}$. Also, it is generated by $(1,0)$ and $(0,1)$. So consider the homomorphism where we send $x$ to $(1,0)$ and $y$ to $(0,1)$. Can you finish the proof from here?