If $\phi: G\rightarrow G'$ be a group onto homomorphism then show that $|G'|$ divides $|G|$

Question

If $\phi: G\rightarrow G'$ be a (finite) group onto homomorphism then show that $|G'|$ divides $|G|$

$\phi: G\rightarrow G'$ be a group onto homomorphism then by Isomorphism theorem,

$G/Ker~ \phi \simeq G'$ and then $|G/Ker~ \phi|= |G'|$ i.e $$|Ker~ \phi|=\frac{|G|}{|G'|} $$

But $|G'|$ divides $|G|$ can be concluded only when $|Ker~ \phi|$ exist finitely. What to do?

Is any other alternative method to solve?

Answer 1

Since the problem asks us to prove $|G'|$ divides $|G|$ we assume that $G$ and $G'$ are already finite groups, which implies that $\ker\phi$ is also finite (as it is a subgroup of $G$)

Answer 2

It is slightly more clear to stick to $|G'| \ |Ker \phi|= |G|$, and not to divide by $|Ker \phi|$.

This clearly shows the divisibility in case $|G|$ is finite. If $G$ is not finite, I'd say the question does not make sense. But if one were to make sense of it then it would be again the equality $|G'| \ |Ker \phi|= |G|$ that is relevant.