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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?

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    $\begingroup$ 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. $\endgroup$
    – Vercassivelaunos
    Jan 16, 2021 at 17:03

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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!

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  • $\begingroup$ Thanks for your comment but It doesn't really say anything more than I said myself... $\endgroup$
    – The Lion King
    Jan 16, 2021 at 17:57
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    $\begingroup$ For $\Leftarrow$, what does my hint tell you about the index of both kernels in $G$? $\endgroup$
    – user299843
    Jan 16, 2021 at 18:01
  • $\begingroup$ 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 $\Rightarrow$? $\endgroup$
    – The Lion King
    Jan 16, 2021 at 18:16

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