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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?

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  • $\begingroup$ 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. $\endgroup$ – Kaj Hansen Oct 10 '16 at 19:02
  • $\begingroup$ If $G$ is a finite group, then $ker \phi \subset G$ is also finite. $\endgroup$ – u1571372 Oct 10 '16 at 19:02
  • $\begingroup$ @DietrichBurde Without using Isomorphism theorem directly. $\endgroup$ – rama_ran Oct 10 '16 at 19:20
  • $\begingroup$ 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. $\endgroup$ – Dietrich Burde Oct 10 '16 at 19:25
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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.

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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 $G$)

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  • $\begingroup$ Ok. Understood. Is any other alternative method to solve? $\endgroup$ – rama_ran Oct 10 '16 at 19:09
  • $\begingroup$ This is probably the simplest way to solve it. $\endgroup$ – Jorge Fernández Hidalgo Oct 10 '16 at 19:15

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