If $G$ is finite abelian group and $f: G \rightarrow G$ is a group homomorphism, how can I show that there exists an $n$ such that $\ker(f^n)$ intersects trivially with $f^n(G)$?
My thought: Since $G$ is finite, then there is $n \in \mathbb{N}$ such that $f^{n+1} = f^n$.
Then, I pick $g$ in $\ker(f^n) \cap f^n(G)$, then $f^n(g)=e_G$ and $g=f^n(h)$ for some $h \in G$.
Since $f$ is a group homomorphism then so $f^n$, it follows that $f^n(e_G) =e_G$.
Can I say that $g = e_G$?
It seems that $\ker(f^n)$ is always the trivial subgroup of $G$.