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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know that a finite group need not have a subgroup for any particular prime index (for example, $A_4$ has no subgroup of index $2$, but it does have one of index $3$). Is there a nontrivial finite group with no subgroups of prime index?

share|cite|improve this question
The trivial group. – RghtHndSd May 21 '14 at 2:23
Thanks, edited. I should have seen that myself... – Nishant May 21 '14 at 2:31
See also for a stronger (but false) claim, and for a closer-to-true claim. Both are answered by the same paper of Guralnick, but might be interesting questions. – Jack Schmidt May 21 '14 at 3:07
up vote 13 down vote accepted

The simple group $\operatorname{PSL}(2,8)$ has no subgroup of prime index. Its maximal subgroups have indices 9, 28, and 36.

In general $\operatorname{PSL}(2,q)$ has order $q(q-1)(q+1)/\gcd(q-1,2)$. Set $d=\gcd(q-1,2)$.

It has (usually) maximal subgroups that are dihedral of order $2(q+1)/d$ and $2(q-1)/d$. It has a maximal subgroup of order $q(q-1)/d$. If $q$ is a power of $r$ (with the power itself prime), then $\operatorname{PSL}(2,r)$ is a (usually) maximal subgroup. It also has (usually) maximal subgroups of orders $12$, $24$, or $60$ ($A_4$, $S_4$, or $A_5$ depending on some congruences on $q$).

For $q=8$, we get the order is 504. It has no subgroup of order 60 by Lagrange. So we check the (possibly) maximal subgroups of order $2(q+1)/d = 18$, $2(q-1)/d = 14$, $q(q-1)/d = 56$, $2(2-1)(2+1)/\gcd(2-1,2) = 6$ (not maximal in this exceptional case), and $24$ (which in this exceptional case is contained in the order 56 group).

These subgroups have index 28, 36, 9, 84, and 21.

The subgroup classification is in Huppert's Endliche Gruppen Kap II. §8, pages 191-214, culminating in Dickson's Hauptsatz 8.26 on page 213 (which is just the more precise version of what I said above).

Most simple groups have no subgroups of even prime power index. The list of exceptions is fairly short and is given in Guralnick (1983). In particular, the cyclic groups of prime order, a few bizarre PSL(n,q), alternating groups of prime power degree, the prime degree Mathieu groups, and PSp(4,3). Every other simple group is a counterexample to the claim that a non-identity group must have some proper subgroup of prime power index. For example the alternating group $A_{6} = \operatorname{PSL}(2,9)$ has maximal subgroup indices 6, 6, 10, 15, and 15.

share|cite|improve this answer
Can you provide a reference or an argument for this? – RghtHndSd May 21 '14 at 2:35
I can write it up. Burnside and Dickson circa 1910 textbooks have a description of all subgroups of PSL(2,q). – Jack Schmidt May 21 '14 at 2:36
Here is one positive result: if $G$ is a nontrivial finite solvable group, then some subgroup of $G$ has prime index. (proof: use induction) – Mikko Korhonen May 21 '14 at 13:37
@MikkoKorhonen: slightly stronger (from @lhf) if $G$ is finite but not perfect, then $G/G'$ has a subgroup of prime index, so $G$ does as well. – Jack Schmidt May 21 '14 at 13:40

Your Answer


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.