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Let $G$ be a group and $H$ a non-normal subgroup of $G$. Please prove that the order of $G$ is divided by a prime number which is less than $[G:H]$.

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marked as duplicate by YACP, Brian Rushton, Eric Stucky, Paramanand Singh, Paul Jan 3 at 3:44

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What've you tried? Just asking questions without showing any self effort is not considered positively here. –  DonAntonio Dec 28 '13 at 17:11
    
I tried to assume that $o(G)$ is divided only by a prime number which isn't less than $[G:H]$. That means that the prime factorization of $H$ are at least $[G:H]$. I couldn't find how to go on :( –  Ran Kashtan Dec 28 '13 at 17:26

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As $H\not\lhd G$ implies $[G:H]>1$, we also have $o(G)>1$ and hence can let $p$ be the smallest prime divisor of $o(G)$. Since $1<[G:H]\mid o(G)$, we need only consider the case $[G:H]=p$. But:

Proposition. Let $G$ be a nontrivial finite group and $p$ the smallest prime dividing $o(G)$. Then any subgroup $H$ of index $[G:H]=p$ is normal in $G$.

Proof. The group $G$ acts by left multiplication on the set of the $p$ left cosets mod $H$, which gives us a homomorphism $\phi\colon G\to S_p$. For any $g\notin H$, $\phi(g)$ is not the identity. Hence $\ker\phi\le H$. On the other hand, for $h\in H$ the permutation $\phi(H)$ leaves $H$ fixed, hence can be viewed as an element of $S_{p-1}$. As such it has order dividing $(p-1)!$. Since the order of $\phi(h)$ divides the order of $h$ and $(p-1)!$ is relatively prime to $o(G)$, we conclude that $\phi(h)$ is the identity, i.e. $H\le \ker \phi$. Thus $H=\ker\phi$ is normal. $_\square$

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@user Hagen von Eitzen likes to get his points the cheap way, and seems to be quite proud about it. –  Igor Rivin Dec 28 '13 at 18:26
    
Can you please explain why we need only consider the case [G:H]=p and ignore the case: [G:H]>p? –  Ran Kashtan Dec 30 '13 at 21:21
    
I meant [G:H]<p :) –  Ran Kashtan Dec 30 '13 at 21:37
    
I got it! :) Thank you! –  Ran Kashtan Dec 30 '13 at 21:41

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