Relation between a group homomorphism and the composition with itself

Question

Let $G$ be a finite group and $\varphi :G \longrightarrow G$ a group homomorphism. To prove:$\DeclareMathOperator{\Ker}{Ker}\DeclareMathOperator{\Im}{Im}$ $$\Ker(\varphi)=\Ker(\varphi^2) \iff \Im(\varphi)=\Im(\varphi^2)$$

I have some tools but I'm not able to combine them effectively. If $x\in \Ker(\varphi)$, then $\varphi(\varphi(x))=\varphi(e_G)=e_G$, hence $\Ker(\varphi) \subseteq \Ker(\varphi^2)$. With the first isomorphism theorem follows $\Im(\varphi) \cong \Im(\varphi^2)$.

Any hints?

Answer 1

Using the first isomorphism theorem is a good step! It gives $\text{Im}(\varphi)\cong G/\text{Ker}(\varphi)$ and $\text{Im}(\varphi^2)\cong G/\text{Ker}(\varphi^2)$. This should make one direction easy, for the other direction you have already made the first important observation. Good luck!