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Let $X$ be a finite group and $\phi : X \longrightarrow X$ an endomorphism. I read somewhere that if $\phi$ is injective, then it is also surjective, and thus bijective. I have seen a lot of proof on this topic, but never in group theory.

I know that $\phi$ being injective means that 2 elements of $X$ can't have the same image. But how does this imply that $\phi$ is surjective ?

Wicowan
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    The group structure is irrelevant here. – Shaun Nov 29 '22 at 22:28
  • It is a general fact that any injective function from a finite set to itself is surjective. I believe when you talk about seeing "a lot of proofs", you mean you know this result. If $\phi$ is a homomorphism then it is in particular a function from $X$ to itself, so it's just a special case. – Mark Nov 29 '22 at 22:28
  • @Mark yes, I didn't know that group or set didn't matter here, saw a lot of proof on ring theory, modules and vector space, but never specifically on group theory, guess that's why – Wicowan Nov 29 '22 at 22:35
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    Regarding the vector space version of this result, the statement and proof are somewhat different, because the theorem is about finite dimensional vector spaces, not finite vector spaces (well, I suppose it could also be about finite vector spaces over finite fields...) – Lee Mosher Nov 29 '22 at 22:47

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