Homomorphisms from $\mathbb Z \rightarrow \mathbb Z_{6}$

Question

Homomorphisms from $\mathbb Z \rightarrow \mathbb Z_{6}$

I have proven that the number of Homomorphisms from $\mathbb Z_6 \rightarrow \mathbb Z=6$

Is the reverse just simply $6$ also?

in $\mathbb Z_6$ we have $1+1+1+1+1+1=6=0$

let $f(1)=a \in \mathbb Z$

$f(1+1+1+1+1+1)=f(0)=0$

$6f(1)=0$

$6a=0 \in \mathbb Z_6$

So is $a=\mathbb Z$ or is $a=0,1,2,3,4,5$

Any explanation on this would be great.

Answer 1

How did you prove that $\left|\operatorname{Hom}(\mathbb{Z}_6,\mathbb{Z})\right| = 6$? We can state that for every $n \in \mathbb{N}$ $\left|\operatorname{Hom}(\mathbb{Z}_n,\mathbb{Z})\right| = 1$ which result to be the trivial homomorphism, since from the First Homomorphism Theorem

$ \mathbb{Z}_n / \operatorname{Ker}(\varphi) \cong \operatorname{Im}(\varphi) < \mathbb{Z}$, but all the subgroup of $\mathbb{Z}$ are of the form $d\mathbb{Z}$ which of course are infinite, and since $|\mathbb{Z}_n / \operatorname{Ker}(\varphi)| < +\infty$ we have $d = 0$ hence $\operatorname{Ker}(\varphi) = \mathbb{Z}_n$

Conversely to count $\left|\operatorname{Hom}(\mathbb{Z},\mathbb{Z}_6)\right|$ we know they are determined once assigned the image of a generator since $\mathbb{Z}$ is a cyclic group.

Because we can choose $\phi(1)$ in $|\mathbb{Z_{6}}| = 6$ ways there are 6 such homomorphisms.