Prove $\alpha:G\rightarrow H $ is a homomorphism (G,H finite) $\implies$ $|\alpha(G)|$ divides $|G|$ and $|H|$

Question

I'm totally stuck here. Obviously $\alpha(G)$ is a subgroup of $H$, so its order divides the order of $H$, but I don't see how to prove that $|\alpha(G)|$ divides $|G|$. I've gotten as far as showing that the only case that matters is when $|\alpha(G)|<|G|$, but then I don't have an argument about how distinct terms in $G$ mapping to the same thing in $H$ would somehow work out to make the image of the group have size that's a factor of the size of the group.

Any hints would be greatly appreciated!

Answer 1

By the first isomorphism theorem, $$\alpha(G)\simeq G/\ker \alpha$$ Thus in particular its size divides $|G|$.

Answer 2

By the first isomporphism theorem, $\alpha(G)\cong G/\operatorname{ker}\alpha$. Thus $|\alpha(G)|||G|$.

Secondly, by Lagrange, since $\alpha(G)$ is a subgroup, $|\alpha(G)|\vert|H|$.