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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.

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If $2^{n-1}-1$ is a prime number, then the answer is no, by Burnside theorem. –  Seirios Jul 28 '12 at 7:04
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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
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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.

Bibliography

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

http://en.wikipedia.org/wiki/List_of_finite_simple_groups

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!

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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

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