Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

4 Answers

up vote 6 down vote accepted

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.

share|improve this answer
    
+1 As simple and succinct as possible. –  DonAntonio Jun 18 '12 at 16:33
add comment

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 $\phi$?

share|improve this answer
add comment

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 $H$.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.