# Showing a normal subgroup contains a subgroup

Let G be a finite group, $H \le G$, and $N\lhd G$. Suppose $|H|$ and $|G :N|$ are relatively prime. Is it true that $H \le N$?

Since $N$ is a normal subgroup, I know that $NH \le G \implies |NH|$ divides $|G|$. Also $|HN|=\frac{|H||N|}{H\cap N}$, so using the formula $|G :N|=\frac{|G|}{|N|}$ and manipulating I get,

$\frac{|G|}{|HN|}=\frac{|G||H\cap N|}{|H||N|}$

but I do not know how to continue(assuming I am on the right track to begin with.). Any help is much appreciated.

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## 4 Answers

Hint: Consider the image $\pi(H)$ of $H$ in the quotient group $G/N.$ If you can show $\pi(H)$ is trivial then $H \subset N.$ Think about what numbers $|\pi(H)|$ divides.

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+1 As simple and succinct as possible. –  DonAntonio Jun 18 '12 at 16:33

Consider the composition $\phi:H\to G/N$ of the inclusion $H\to G$ with the projection $G\to G/N$. The orders of $H$ and of $G/N$ are coprime: what are the possibe images of $\phi$?

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Let $\varphi:G\to G/N$ be the quotient map. Suppose that $h \in H$; what do you know about the possible orders of $\varphi(h)$? You have information both from the index of $N$ and from the order of $H$.

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The answers already given are perfectly fine, of course, and it is always a good idea to think in terms of homomorphisms, quotients, kernels etc. In this particular case though, I think you were just a step from solving it, so there is no need to change your strategy.

As you observed, the LHS is an integer. Now write the RHS as $\frac{|G:N||H \cap N|}{|H|}$.

Since $\gcd(|H|,|G:N|)=1$, $|H|$ divides $|H \cap N|$. But $H \cap N$ is a subgroup of $H$, so $|H \cap N|$ divides $|H|$. Hence $|H|=|H \cap N|$ which yields $H=H \cap N$. This is obviously equivalent to what you wanted to prove.

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