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.