Let $G$ be a finite simple group and let $H\subset G$ be a abelian subgroup of index $|G:H|=p$ for $p$ some prime. Prove that $H = \{e\}$.

I don't know how to start... Plz someone help me.


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    $\begingroup$ Is $p$ a prime number? $\endgroup$ – Bernard Mar 17 at 12:38
  • $\begingroup$ yes, $p$ is prime number $\endgroup$ – Silement Mar 17 at 13:04
  • $\begingroup$ Hint : show that $H$ such that $|G:H|=p$ is a normal subgroup of $G$. $\endgroup$ – TheSilverDoe Mar 17 at 17:01
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    $\begingroup$ @TheSilverDoe That is not correct in that generality. $\endgroup$ – Tobias Kildetoft Mar 18 at 13:06
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    $\begingroup$ @TobiasKildetoft Sorry, I thought (I don't know why) that $p$ was the smallest prime number dividing the order of $G$... In this case, I think that $H$ is normal. In general, you are right, it is not the case. Thanks you for the correction ! $\endgroup$ – TheSilverDoe Mar 18 at 15:01

I can offer two proofs of this statement. The first is short, but makes use of a difficult theorem (due to Herstein; an alternative proof on MathOverflow):

Theorem: A finite group with an abelian maximal subgroup is solvable.

The proof is not elementary, as it involves the Frobenius complement. But it makes short work of the present question:

In our case, $H$ is a maximal subgroup of $G$ because $|G:H|$ is prime; thus $G$ is solvable. However, the only simple groups which are solvable are the prime-order cyclic groups. Hence $G\cong C_p$ and $H$ is trivial.

By making earlier use of the hypothesis that $G$ is simple and borrowing an idea from Herstein, I have cobbled together an elementary proof:

If $G$ is abelian, then it must be prime-order cyclic and we are done. So we will assume that $G$ is a nonabelian simple group and show that this results in a contradiction.

$H$ is not trivial, for then $G$ would be of prime order and thus abelian. Thus $H$ is not normal. Let $K$ be a conjugate of $H$. Since $K\cong H$, $K$ is also abelian.

Claim: $H\cap K$ is trivial.

If $h \in H\cap K$, then the centralizer of $h$ contains $H$ (since $H$ is abelian) and $K$ (likewise). Since $H$ is maximal and the centralizer of $h$ is larger, it must be all of $G$. But $G$ is nonabelian simple, so its center is trivial. Hence $h = 1$.

Let $n = |G|$, so $|H| = |K| = n/p$. Then $|HK|=\dfrac{|H|\,|K|}{|H\cap K|} = n^2/p^2$. $HK$ is a subset of $G$, so $n^2/p^2 \le n$; thus $n \le p^2$. It cannot be the case that $n=p^2$, since all groups of order $p^2$ are abelian. Hence $n \lt p^2$ and $n/p \lt p$. The number of Sylow $p$-subgroups of $G$ is a divisor of $n/p$ and is congruent to $1\pmod p$, so it must be $1$. Thus the Sylow $p$-subgroup is normal, which contradicts the simplicity of $G$.

I suspect that there is a simpler elementary proof, so I hope that this awkward approach will provoke someone into embarrassing me by posting it.

  • $\begingroup$ Shouldn't the end of your proof be "hence the $p$-Sylow is normal; therefore it is $G$, which is absurd as nonabelian $p$-groups aren't simple" ? Otherwise where did you remove the possibility that $G$ be a $p$-group ? $\endgroup$ – Max Mar 19 at 11:23
  • $\begingroup$ @Max The group is already of order less than $p^2$ and not cyclic of order $p$, so it cannot be a $p$-group. $\endgroup$ – jgon Mar 19 at 12:56
  • $\begingroup$ @jgon : oh right, my bad ! $\endgroup$ – Max Mar 19 at 13:18

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