We know that the number of homomorphisms from $\mathbb{Z}_n$ to $\mathbb{Z}_m$ is equal to $(n,m)$, the greatest common divisor of $n$ and $m$. Also, the number of homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}_m$ is going to $m$.
My question is, how many homomorphisms are there from $\mathbb{Z}_n \times \mathbb{Z}$ to $\mathbb{Z}_m$?
We have $$\mathbb{Z}_n \times \mathbb{Z} = \langle (1,0), (0,1): n \cdot (1,0) = 0\rangle,$$ so determining a homomorphism can be done by specifying the image of the generators. Hence we need to send $(1,0)$ to an element that satisfies $n \cdot \phi(1,0) = 0$ in $\mathbb{Z}_m$.
Again there are $(n,m)$ ways to do this, moreover we need to send $(0,1)$ to something, and as mentioned sending this to any element of $\mathbb{Z}_m$ will define a homomorphism.
Can we then say the number of homomorphisms from $\mathbb{Z}_n \times \mathbb{Z}$ to $\mathbb{Z}_m$ is $(n,m)\cdot m$?
Best Answer
If $K$ is an Abelian group and there are $a$ homomorphisms $G\to K$ and $b$ homomorphisms $H\to K$ then the number $c$ of homomorphism $G\times H\to K$ is $ab$. Indeed, if $\phi_1:G\to K$, $\phi_2:H\to K$ are homomorphisms then $\phi: (g,h)\mapsto \phi_1(g)\phi_2(h)$ is a homomorphism $G\times H\to K$ since $K$ is Abelian. Its restrictions to $G\times 1$ and $1\times H$ are $\phi_1,\phi_2$. Hence $c\ge ab$.
On the other hand for every homomorphism $\phi$ from $G\times H\to K$, its restrictions to $G\times 1$ and $1\times H$ are homomorphisms $\phi_1,\phi_2$ such that $\phi(g,h)=\phi_1(g,1)\phi_2(1,h)$. So $c\le ab$.
Thus $c=ab$.