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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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    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 $\{1\}$.2012-03-30
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    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.2012-03-30
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    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.2012-03-30
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    And what is the order of $G/H$?2012-03-30
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    It must be $m$. I think I got this, but my mind is clouded with some (personal life) problems right now.2012-03-30
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    I mean, the order of $\pi(H')$ is $n$ minus the order of the intersection of $H$ and $H'$ (which is the kernel of $\pi$ when it is restricted to $H'$. If that's a divisor of $n$ then the exercise is over... but, must it be a divisor of $n$?2012-03-30
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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