Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It is well known that in finite solvable groups, any maximal subgroup has prime power index.

Question: If $G$ is a finite group in which any maximal subgroup has prime power index, is $G$ solvable?

Up to groups of order $<180$, the answer to the question seems to be yes.

share|improve this question
    
You seem to have missed one of the few, or the only one, known counter examples of order 168, @Marshal –  DonAntonio Mar 1 at 9:34

1 Answer 1

The answer is no. A counterexample is given by the simple group of order $168$ (every maximal subgroup has index $8$ or $7$). Using the classification of finite simple groups, it has been proven by Guralnick [1] that this is the only example of a finite nonabelian simple group with this property.

There is a partial converse due to Philip Hall, which states the following

Theorem: Suppose that $G$ is a finite group and that for any maximal subgroup $M$, the index $[G:M]$ is prime or a square of a prime. Then $G$ is solvable.

A proof can be found in Endliche Gruppen by Huppert, pg. 718, Satz VI.9.4. Also in A Course on Group Theory by Rose, pg. 281, Theorem 11.16 and The Theory of Groups by M. Hall, pg. 161, Theorem 10.5.7.

There is also the following result due to Huppert (1954):

Theorem: A finite group $G$ is supersolvable if and only if every maximal subgroup of $G$ has prime index.

Proof: Suppose that $G$ is supersolvable and let $M$ be a maximal subgroup. Now there exists a normal subgroup $P$ of prime order. If $P \leq M$, then $M/P$ is a maximal subgroup of the supersolvable group $G/P$, and the result follows by induction. Otherwise $P \cap M = 1$, and in this case $G = PM$ since $M$ is maximal. In particular $[G:M] = |P|$.

For the other direction, a proof is in Endliche Gruppen by Huppert, pg. 718, Satz VI.9.5, and also in Group Theory by Scott, pg. 226, Theorem 9.3.8 and The Theory of Groups by M. Hall, pg. 162, Theorem 10.5.8.

[1] R. Guralnick, Subgroups of prime power index in a simple group, Journal of Algebra, Volume 81, Issue 2, April 1983, Pg. 304–311 DOI

share|improve this answer
    
Unless its proof is too long and complicated,Can you give proof or some refference. –  mesel Mar 1 at 9:09
    
Thansk for edit. –  mesel Mar 1 at 9:11
    
In his book "A course in Group Theory" (page 279,immediately after corollary 11.15), John S. Rose mentions that precisely that simple group of order $\;168\;,\;\;GL_3(\Bbb Z_2)\;$ , is the only known example of non-soluble group all of which maximal subgroups have index a power of a prime (as mentioned above, $\;7\,,\,8\;$). The book's from 1978, a little before Guralnick's paper. –  DonAntonio Mar 1 at 9:22
    
@DonAntonio: Right, that's how I first found out about this example. Also, I think you mean non-soluble simple group. –  Mikko Korhonen Mar 1 at 9:29
3  
@DonAntonio: I think something like $\operatorname{GL}_3(\mathbb{F}_2) \times C_p$, with $p > 7$ prime should be an example. –  Mikko Korhonen Mar 1 at 9:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.