4
$\begingroup$

I need help showing this result:

"Let $G$ be a group such that $|G|=nm$ where $m$ and $n$ are relatively prime. Suppose that there exists a normal subgroup H of G such that $|H|=n$. Show that $H$ is the only subgroup with order $n$."

Can someone give me a light?

$\endgroup$
9
$\begingroup$

Consider the order of the image of any alleged subgroup of order $n$ under the canonical map from $G$ to $G/H$.

$\endgroup$
  • $\begingroup$ Can I say that $|\pi (H')|=n$ ($\pi$ is the canonical map)? Because I think I should think about $ker (\pi)$ which, when restricted to $H'$ (where $H'$ is the group other than $H$ with order $n$) as being the intersection of $H$ and $H'$ which is not necessarily $\{1\}$. $\endgroup$ – Marra Mar 30 '12 at 2:13
  • $\begingroup$ If your notation means what I think it means, then $\pi(H')$ is a subgroup of $G/H$, and that has implications for its order. $\endgroup$ – Gerry Myerson Mar 30 '12 at 2:59
  • $\begingroup$ Yes, then it must divide the order of $G/H$, but that does not mean that it must divide m or n because they are not prime numbers themselves. $\endgroup$ – Marra Mar 30 '12 at 9:45
  • $\begingroup$ And what is the order of $G/H$? $\endgroup$ – Gerry Myerson Mar 30 '12 at 11:25
  • $\begingroup$ It must be $m$. I think I got this, but my mind is clouded with some (personal life) problems right now. $\endgroup$ – Marra Mar 30 '12 at 14:05

Your Answer

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

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