# Let $G$ be a finite simple group and let $H\subset G$ be a abelian subgroup of index $|G:H|=p$. Prove that $H = \{e\}$. [closed]

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.

## closed as off-topic by Derek Holt, jgon, mrtaurho, Cesareo, Thomas ShelbyMar 21 at 11:10

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• Is $p$ a prime number? – Bernard Mar 17 at 12:38
• yes, $p$ is prime number – Silement Mar 17 at 13:04
• Hint : show that $H$ such that $|G:H|=p$ is a normal subgroup of $G$. – TheSilverDoe Mar 17 at 17:01
• @TheSilverDoe That is not correct in that generality. – Tobias Kildetoft Mar 18 at 13:06
• @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 ! – 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.

• 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 ? – Max Mar 19 at 11:23
• @Max The group is already of order less than $p^2$ and not cyclic of order $p$, so it cannot be a $p$-group. – jgon Mar 19 at 12:56
• @jgon : oh right, my bad ! – Max Mar 19 at 13:18