Showing $\mathbb{Z}_6$ is an injective module over itself I want to show that $\mathbb{Z_{6}}$ is an injective module over itself. I was thinking in using Baer's criterion but not sure how to apply it. So it suffices to look at non-trivial ideals, the non-trivial ideals of $\mathbb{Z_{6}}$ are:
(1) $I=\{0,3\}$ 
(2) $J=\{0,2,4\}$
So take a $\mathbb{Z_{6}}$-map $f: I \rightarrow \mathbb{Z_{6}}$. Since $f$ is a group homomorphism it must map generators to generators right? so $3 \mapsto 1$ and $0 \rightarrow 0$. Now can we say suppose $f(1)=k$ then define $g: \mathbb{Z_{6}} \rightarrow \mathbb{Z_{6}}$ by sending the remaining elements, (those distinct from 0 and 3), say n, to $nk$?

I found a solution for the 'general' case:
Let I be a ideal of $\mathbb{Z}/n\mathbb{Z}$ then we know that $I=\langle \overline{k} \rangle$ for some $k$ such that $k\mid n$.
If $f:I\rightarrow \mathbb{Z}/n \mathbb{Z}$ is a $\mathbb{Z}/n \mathbb{Z}$-morphism then $im f\subset I$. To show this we note that if $\overline{x}\in im f$ then there exist $\overline{c}=\overline{lk}$ with $\overline{l}\in\mathbb{Z}/n \mathbb{Z}$ s.t. $f(\overline{c})=\overline{x}$, but $n=ks$ for some $s$, so $\overline{0}=f(\overline{ln})=\overline{s}\cdot f(\overline{lk})=\overline{sx}$, then $sx=nt$ for some $t$, but again since $n=ks$, we get $x=tk$.
In particular for $\overline{k}\in I$ we have $f(\overline{k})=b\overline{k}$ for some $b$. So for any $x\in I$ we have $f(x)=bx$ then we just take the extension of $f$ to be the map $\mathbb{Z}/n\mathbb{Z}\rightarrow \mathbb{Z}/n\mathbb{Z}$ where $x\mapsto bx$.
