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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?

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up vote 9 down vote accepted

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

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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 $\{1\}$. – Marra Mar 30 '12 at 2:13
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. – Gerry Myerson Mar 30 '12 at 2:59
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. – Marra Mar 30 '12 at 9:45
And what is the order of $G/H$? – Gerry Myerson Mar 30 '12 at 11:25
It must be $m$. I think I got this, but my mind is clouded with some (personal life) problems right now. – Marra Mar 30 '12 at 14:05

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