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

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?

• The statement only makes sense for finite groups $G$. Since $\ker(\phi)$ is a subgroup of $G$, any situations where the kernel is infinite are irrelevant. Oct 10, 2016 at 19:02
• If $G$ is a finite group, then $ker \phi \subset G$ is also finite. Oct 10, 2016 at 19:02
• @DietrichBurde Without using Isomorphism theorem directly. Oct 10, 2016 at 19:20
• We could just show that the cardinalities $|G/\ker(\phi)|$ and $|G'|$ coincide. This is "less" than the "isomorphism theorem". But the proof would be suspiciously similar to the proof of the isomorphism theorem. So no real progress. Perhaps one would ask why you don't want the isomorphism theorem; it is so basic. Oct 10, 2016 at 19:25

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.
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 $G$)