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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?

1 Answers 1

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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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    You're right to consider the kernel of $\pi$ when it is restricted to $H'$; let's call that kernel $K$. Then $\pi(H')$ is isomorphic to $H'/K$, which tells you what you need to know about its order.2012-03-30