Surjective Homomorphism $f : \mathbb{Z}/12\mathbb{Z}\times \mathbb{Z}/6\mathbb{Z}\longrightarrow\mathbb{Z}/6\mathbb{Z}$.

Question

Good morning everyone,

I'd like to discuss the following exercise.

In general when I have to find the number of homomorphisms $g: \mathbb{Z}/m\mathbb{Z}\longrightarrow\mathbb{Z}/n\mathbb{Z}$, I know that is $(m,n)$ , but I have also have to consider the condition that $\operatorname{ord}(f(x)) \mid \operatorname{ord}(x)$.

So in this case I'd say that $\operatorname{ord}(f(x)) \mid \operatorname{ord}(x)$ always, because $\operatorname{ord}(f(x))$ is at most six, so I'd say that the total number of homomorphisms between these two groups is $72$.

I also feel that this conclusion is completely wrong, but I'd like to notice where.

The second point asked me to count the number of surjective homomorphisms: in general I would find a generic $f(x_n) \in \mathbb{Z}/6\mathbb{Z}$ with $x_n \in \mathbb{Z}/12\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$ such that $f(x_n)$ has exact order $6$.

Could you help with my doubts and try to fix my misunderstandings ?

Thanks.

Answer 1

Hint

Let denote $\mathbb Z_n:=\mathbb Z/n\mathbb Z$. Since $\mathbb Z_6=\left<1\right>$, all homomorphism are determinated by elements $k\in \mathbb Z_{12}\times \mathbb Z_6$ s.t. $f(k)=1$. So surjective homomorphism are determinated by elements $k$ of order $6$ in $\mathbb Z_{12}\times \mathbb Z_6$.