# Relation between a group homomorphism and the composition with itself

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?

• If $\ker\varphi=\ker\varphi^{2}$ then that means that $\varphi$ fixes the kernel, i.e. $\varphi(\ker\varphi)=\ker\varphi$. This then means that the image of a coset of the kernel is also a coset of the kernel. Also note that different cosets cannot be mapped to the same coset. Jan 16, 2021 at 17:03

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!
• For $\Leftarrow$, what does my hint tell you about the index of both kernels in $G$? Jan 16, 2021 at 18:01
• They have the same cardinality and since $G$ is finite and one inclusion always holds, then they are equal. That's a little step that I was missing, thank you. Any similar hints on how to prove $\Rightarrow$? Jan 16, 2021 at 18:16