If $\phi:G \to \overline G$ is group homomorphism, then prove that $|\phi(G)|$ divides $|G|$.

Question

Is there any other short proof of it? Thanks.

I don't know of a short alternate proof. I could use the fundamental theorem, but that's really using the same idea. — Michael Burr, Jul 01 '18 at 13:31
Easy. Prove that $\phi(G)$ is a subgroup of the target group. Then use Lagrange's Theorem. — the_fox, Jul 01 '18 at 13:31
@the_fox That only shows that $|\phi(G)|$ divides $|\overline{G}|$, not $|G|$. — Michael Burr, Jul 01 '18 at 13:32
Yeah, I think I agree with Michael Burr. I don't see any way to prove that without invoking (directly or indirectly) the First Isomorphism Theorem. — the_fox, Jul 01 '18 at 13:35
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 — Maxime Ramzi, Jul 01 '18 at 13:37
@Max You don't need anything other than the First Isom. Thm. — the_fox, Jul 01 '18 at 13:39

lhf · Answer 1 · 2018-07-02T01:03:36.133

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 $b\in \overline G$:

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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