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.

I would like to know: Is there any simple group of order $2^{n}(2^{n}-2)$? This is a special case of a group of order $p^{2}-1$ when $p$ is Mersenne prime. Thanks.

share|improve this question
If $2^{n-1}-1$ is a prime number, then the answer is no, by Burnside theorem. –  Seirios Jul 28 '12 at 7:04
mathoverflow.net/questions/103386/… is the (less general, I believe) question where one also has a hypothesis on the automorphism group of $G$ –  Jack Schmidt Jul 28 '12 at 16:11
Note that there is a simple group of order $13^{2}-1$ and a simple group of order $19^{2}-1,$ so you do need some condition on a prime $p$ to exclude simple groups of order $p^{2}-1.$ –  Geoff Robinson Jul 29 '12 at 15:40

2 Answers 2

up vote 3 down vote accepted

For a simple numerical invariant, one can use that a known, non-abelian, finite, simple group $G$ always satisfies $|U|^2 < |G|$ when $U$ is a Sylow $p$-subgroup of $G$. This was shown in (Kimmerle, et al., 1990, Theorem 3.6) and (Mazurov–Zenkov, 1995, Corollary 1).

However, if $p=2^n-1$ is prime, $|G|=p^2-1 = 2^n(2^n-2) = 2^{n+1}(2^{n-1}-1)$,and $U$ is a Sylow 2-subgroup, then $|U|^2 = 2^{2n+2} > 2^{2n} > |G|$.

Proof sketch

Here is an intuitive argument why a Sylow $p$-subgroup $U$ should satisfy $|U|^2 \leq |G|$ when $O_p(G)=1$: Consider $|U|^2 = |U| |U^g| = |U U^g | |U \cap U^g| \leq |G| |U \cap U^g|$. If $|U|^2 > |G|$, then $|U \cap U^g | > 1$. In most groups, $O_p(G) = U \cap U^g$ for some $g$, so we'd have a non-trivial $p$-core. Of course in some groups it requires 3 Sylow $p$-subgroups to get the $p$-core, so this is not a proof in general. However, (Mazurov–Zenkov, 1995) uses induction and a result on defect groups of blocks (and a small amount of case by case analysis) to show that in a known, non-abelian, finite, simple group, the $p$-core is always the intersection of two Sylow $p$-subgroups.


  • Kimmerle, Wolfgang; Lyons, Richard; Sandling, Robert; Teague, David N. “Composition factors from the group ring and Artin's theorem on orders of simple groups.” Proc. London Math. Soc. (3) 60 (1990), no. 1, 89–122. MR1023806 DOI:10.1112/plms/s3-60.1.89
  • Mazurov, V. D.; Zenkov, V. I. “Intersections of Sylow subgroups in finite groups.” The atlas of finite groups: ten years on (Birmingham, 1995), 191–197, London Math. Soc. Lecture Note Ser., 249, Cambridge Univ. Press, Cambridge, 1998. MR1647422 DOI:10.1017/CBO9780511565830.019
share|improve this answer

You should've thought of just going through the list and see if you could find one with your desired order! Here it is :


Unless you didn't know that finite simple groups were already fully classified up to isomorphism. =)

But since you speak about Mersenne Primes, don't bother about going through the list if you want to assume that $2^{n-1} - 1$ is prime, because Burnside's theorem says that you won't find any such group (Seirios did the comment first, but I must admit any group theorist who has once seen Burnside's theorem remembers it for the rest of his life everytime someone speaks of the order of a group! It is a must-see if you haven't yet.)

Hope that helps!

share|improve this answer
Thank you so much. Yes when $2^{n-1}-1$ is prime there is not any simple group, but I would like to know whether there is a simple group when $2^{n-1}-1$ is not prime. –  S. T Jul 28 '12 at 7:28

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.