0
$\begingroup$

I am studying Dummit & Foote abstract algebra. This question is what I have a problem.

Let $G$ be a finite group, $H$ be a subgroup of $G$, $N$ be a normal subgroup of $G$. If $|H|$ and $|G:N|$ are relatively prime, prove that $H$ is a subgroup of $N$.

I founded that $H$ is a subset of $N$. But I can't find how to prove about subgroup. Does normality implies that subset becomes a subgroup? If not, how can I prove it?

$\endgroup$
2
  • $\begingroup$ Doesn't it suffice to show that $H \subset N$? Since $H$ is a subgroup of $G$, it is itself a group under the operation of $G$, or $N$. ($H \subset N \leq G$). Thus $H$ is a subgroup of $N$. $\endgroup$
    – luxerhia
    Aug 7, 2020 at 8:23
  • $\begingroup$ But is it not one of the data that $H$ is a (sub)group? $\endgroup$
    – drhab
    Aug 7, 2020 at 8:23

1 Answer 1

1
$\begingroup$

Your hypothesis includes that H is a subgroup of G, to prove that it is a subgroup of N, you only need to prove that it is a subset of N which you just did.

$\endgroup$
1
  • $\begingroup$ Ah, it was a stupid question. Thanks for your help. $\endgroup$
    – user814479
    Aug 7, 2020 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy