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Let $G$ be a finite group and $H$ a subgroup of index $p$, where $p$ is a prime. If $\operatorname{gcd}(|H|, p-1)=1$, then $H$ must be normal. Does somebody have a quick proof of this?

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I've never seen this. The generalization I have seen is: If H is a subgroup of index p, where p is the smallest prime dividing |G|, then H is normal. planetmath.org/… –  Graphth Apr 14 '11 at 22:44
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@Numth: Then this would be a corollary of that, since if $p$ is the smallest prime dividing $|G|$, no smaller prime can divide $|H|$, and thus $\gcd(|h|,p-1)=1$. –  joriki Apr 14 '11 at 22:48
    
@joriki I know. I just provided a partial answer. –  Graphth Apr 14 '11 at 22:52
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here's one generalization (not the one you want): if $[G:H]=p$ and $p$ is the minimal prime divisor of $G$ then $H$ is normal ( groupprops.subwiki.org/wiki/… ) –  yoyo Apr 14 '11 at 22:57
    
@yoyo That's what I already said. –  Graphth Apr 15 '11 at 0:29
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2 Answers

up vote 7 down vote accepted

By induction on $|G|$, we can assume that there is no non-trivial normal subgroup $N$ of $G$ contained in $H$. Thus the action of $G$ on the conjugates of $H$ gives an embedding of $G$ into $S_p$, of order $p!$. Thus $|H|$ divides $(p-1)!$, and so $p$ does not divide $|H|$. Thus the Sylow p-groups of $G$ are cyclic of order $p$.

Now if any $h\in H$ normalized a Sylow p-group $P$, then $h$ would map into $Aut(P)\cong C_{p-1}$, and by hypothesis the image would be trivial. That is, any $h\in H$ normalizing $P$ also centralizes $P$. It follows that $N_G(P)=C_G(P)$, and by the Burnside Transfer theorem, $G$ has a normal subgroup $M$ of index $p$. Of course, any Sylow q-group of $H$ is then contained in $M$, and so $H$ is contained in $M$; that is, $H=M$ is normal.

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Should probably mention there is a slightly faster way to do this, using Frobenius's theorem on normal p-complements; basically, since we can assume $|G|=pm$ where $(m,p-1)=1$, we get a normal p-complement $M$ in $G$ (see for example Corollary 9.22 in Isaacs's Algebra), and then we show $H=M$ just as I did in the end. –  user641 Apr 15 '11 at 0:12
    
Another short proof: Suppose note, and let $H_1,\ldots,H_p$ be the conjugates of $H$; then $G$ acts on these as a transitive group of degree $p$, and since by hypothesis $G$ doesn't live in $AGL(1,p)$, $G$ must be 2-transitive; that is, $(p-1)$ divides the order of $G_1=H_1$, contradiction. This proof shows something stronger: if $H$ is not normal, then either $|H|$ divides $(p-1)$ or vice versa. –  user641 Apr 15 '11 at 23:59
    
Sorry, in my "stronger" statement above, I was thinking of the core-free case. –  user641 Apr 16 '11 at 0:11
    
It might help to put the extra line $N_G(P) = PN_H(P)$ in your proof. –  Geoff Robinson Jul 10 '11 at 9:41
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Note: As pointed out in the comments this answer is missing a crucial detail, and unfortunately I have no idea if it can be salvaged.

This is a slightly more elementary way to show this:

Assume $H$ is not normal. Then clearly $N_G(H) = H$ and $G$ acts on the $|G:H| = p$ conjugates of $H$ by conjugation. The stabilizer of $H$ under this action is just $H$, which then acts on the other $p-1$ conjugates.

The orbit of a $gHg^{-1}$ under this action consists of all $g'Hg'^{-1}$ with $g'\in Hg$ and thus all the orbits have the same size. On the other hand, their sizes all divide the order of $H$, which is a contradiction since $gcd(|H|,p-1) = 1$

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Why do all orbits have the same size? That is, why does $ | H^g\cap H | = | H^{g^2}\cap H | $? –  user641 Apr 15 '11 at 23:23
    
Tobias, I agree with Steve D. your proof is false. Aren't you mixing up the action of G on the conjugates of H and the action of H on these conjugates? –  Nicky Hekster Apr 22 '11 at 8:03
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