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If $\phi:G \to \overline G$ is group homomorphism, then prove that $|\phi(G)|$ divides $|G|$, where $|G|$ is finite. One way to prove this is this: we know that $\phi:G \to \phi(G)$ is an $n$-to-$1$ mapping, where $n=\lvert\ker \phi\rvert$. So $|\phi (G)|\cdot n=|G|$.

Is there any other short proof of it? Thanks.

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  • $\begingroup$ (Provided $|G|$ is finite). $\endgroup$ – Michael Burr Jul 1 '18 at 13:29
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    $\begingroup$ I don't know of a short alternate proof. I could use the fundamental theorem, but that's really using the same idea. $\endgroup$ – Michael Burr Jul 1 '18 at 13:31
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    $\begingroup$ @the_fox That only shows that $|\phi(G)|$ divides $|\overline{G}|$, not $|G|$. $\endgroup$ – Michael Burr Jul 1 '18 at 13:32
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    $\begingroup$ You could use the proof of Lagrange's theorem to show that $|Ker\phi||G/Ker\phi| = |G|$ and the first isomorphism theorem to show $G/Ker\phi \simeq \phi(G)$ but it's essentially the same thing and I don't know if it's shorter $\endgroup$ – Maxime Ramzi Jul 1 '18 at 13:37
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    $\begingroup$ @Max You don't need anything other than the First Isom. Thm. $\endgroup$ – the_fox Jul 1 '18 at 13:39
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Your proof is the shortest I know. Here it is again with more detail:

Consider the fibers of $\phi$ given by $\phi^{-1}(b)$ for $b\in \overline G$:

  • $b \notin \phi(G) \implies \phi^{-1}(b)=\emptyset$.

  • $b \in \phi(G) \implies b=\phi(a), \ a \in G \implies \phi^{-1}(b)=a\ker\phi$.

Therefore, all fibers have the same number of elements. The claim follows because the number of fibers is exactly $|\phi(G)|$.

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