How can I prove that there are no finite groups $G$ where there exists an endomorphism $G\rightarrow G$ that is injective but not surjective, or other way round?
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2This has nothing to do with groups. The statement is true for all maps from a finite set to itself. – Derek Holt Mar 23 '15 at 19:30
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If $X$ is a finite set and $f:X \to X$ is any function, then the conditions below are equivalent:
$f$ is bijective
$f$ is injective
$f$ is surjective
lhf
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See http://math.stackexchange.com/questions/989043/rigorous-proof-that-surjectivity-implies-injectivity-for-finite-sets for proofs. – lhf Mar 23 '15 at 19:30
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It is not a problem about groups but a general fact about sets:
Let $E, F$ be finite sets such that $\lvert E\rvert=\lvert F\rvert$, and $f\colon E\rightarrow F$. The following are equivalent:
$f$ is injective;
$f$ is surjective;
$f$ is bijective.
(1) $\Rightarrow$ (2): If $f$ is injective, $\lvert f(E)\rvert=\lvert E\rvert=\lvert F\rvert$, hence $f(E)=F$, which means $f$ is surjective. (2) $\Rightarrow$ (1) by contraposition: if $f$ is not injective, it can't be surjective since $\lvert f(E)\rvert<\lvert E\rvert=\lvert F\rvert$.
Since (1) $\Leftrightarrow$ (2), it is trivial both are equivalent to (3).
