Let $G$ be a finite group, $|G|=n$, there are $x\in \mathbb{N}$ conjugacy classes. Prove there are $xn$ homomorphisms from $\mathbb{Z}^2$ to $G$.

Question

Let $G$ be a finite group, $|G|=n$, there are $x\in \mathbb{N}$ conjugacy classes. Prove there are $xn$ homomorphisms from $\mathbb{Z}^2$ to $G$.

My approach is to find out the numbers of the normal subgroups of $\mathbb {Z}^2$ and using "Every normal subgroup is the kernel of some homomorphism", is it correct?

I have no idea how to find out it, any help is welcome.

Answer 1

Let $\phi:\mathbb{Z}^2\rightarrow G$ be a homomorphism. For $(x,y)\in \mathbb{Z}^2$, note that $\phi(x,y)=\phi(x(1,0)+y(0,1))=x\phi(1,0)+y\phi(0,1)$. Hence the homomorphism is determined by the image of $(1,0)$ and $(0,1)$ under $\phi$.

Next, observe that $\phi(1,0)\phi(0,1)=\phi((1,0)+(0,1))=\phi((0,1)+(1,0))=\phi(0,1)\phi(1,0)$. Therefore the number of homomorphisms from $\mathbb{Z}^2$ to $G$ is the same as the number of commuting pairs in $G$ which is $xn$. The proof of this result can be found in here.