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My question is very simple and maybe trivial. Here it is.

Is the index of a maximal subgroup in a finite group always a prime number?

Thanks in advance.

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4 Answers 4

up vote 17 down vote accepted

If a group has the property that all its maximal subgroups have prime index, then it is solvable; this follows from a theorem of P. Hall.

In fact, Hall's theorem reaches that conclusion if the indices are prime or squares of primes. A later theorem of Huppert states that if all of them are prime, the group is supersolvable.

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Thomas, some additional trivial observations: your statement is correct for normal maximal subgroups. (However, not true for maximal normal (e.g. the identity subgroup is maximal normal in any non-abelian simple group)). And, if every maximal subgroup is normal, this is equivalent to G being nilpotent.

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No. The smallest counter-example I can think of is $A_5$, the alternating group on $5$ elements which has order $60$, which has a maximal subgroup of order $10$ and thus index $6$ in $A_5$ (more information can be found here).

However, this does hold for finite abelian groups. To see this, use the fundamental theorem of finite abelian groups to write out the abelian group $G$ as $\mathbb{Z}_{p_1^{e_1}}\oplus \mathbb{Z}_{p_2^{e_2}}\oplus \cdots \oplus \mathbb{Z}_{p_n^{e_n}}$. Each maximal subgroup of $G$ contains all of all but one group and a maximal subgroup of $\mathbb{Z}_{p_i^{e_i}}$ for some $i$. But the maximal subgroups of $\mathbb{Z}_{p_i^{e_i}}$ are of order $p_i^{e_i - 1}$, so the index of them in $\mathbb{Z}_{p_i^{e_i}}$ is $p_i$ and thus the index of the maximal subgroup of $G$ is $p_i$ as well.

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1  
That is the smallest counterexample, in fact :) –  Mariano Suárez-Alvarez Apr 27 '11 at 14:42
    
@Mariano: I realized that when I saw your answer concerning solvability. Which is very interesting, by the way. –  Alex Becker Apr 27 '11 at 14:45
7  
The smallest counterexample is A4, not A5. The Sylow 3-subgroup has index 4, and 4 isn't prime. –  Jack Schmidt Apr 27 '11 at 17:19
    
@Jack: Oh, thanks. I misread Mariano's answer as giving its converse as well. –  Alex Becker Apr 27 '11 at 21:28

It may be worth remarking that in a finite solvable group $G$, all maximal subgroups have prime power index (there are already examples given in comments of solvable groups with maximal subgroups not of prime index). I am not sure who was the first person to prove this, but it is well-known, and can be found, for example, in D. Gorenstein's book "Finite Groups" (its proof is elementary compared to many in that book). The proof is as follows: let $M$ be a maximal subgroup of $G$. We may suppose that $M$ contains no nonidentity normal subgroup of $G$ (if $K$ is one such, work in $G/K$). Let $N$ be a minimal normal subgroup of $G$. Then $N$ is an elementary Abelian $q$-group for some prime $q$. Now $N$ is not contained in $M$, so $NM$ is a subgroup of $G$ which strictly contains $M$. Hence $NM = G$. Then $[G:M] = [N:N \cap M]$, and this is a power of $q$.

Note also that if $G$ is a finite solvable group with $\Phi(G) = 1$ and $L$ is any minimal normal subgroup of $G$, then there is a maximal subgroup $H$ of $G$ with $[G:H] = |L|$ (this is again well-known). Recall that the Frattini subgroup $\Phi(G)$ is the intersection of all maximal subgroups of $G$, and is normal. For again, $L$ is an elementary Abelian $q$-group for some prime $q$. Since $\Phi(G) = 1$, there is a maximal subgroup $H$ of $G$ which does not contain $L$. Then $G = LH$. Furthermore, $L \cap H$ is normal in $G$ (since $L$ is Abelian, $L \cap H \lhd L$, and since $L \lhd G$, we also have $L \cap H \lhd H$). But $L$ is not contained in $H$, so $L \cap H$ is a proper subgroup of $L$. By the minimality of $L$ as a normal subgroup of $G$, we must have $L \cap H = 1$, so that $[G:H] = |L|$.

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