A finite group $B$ is said to be a B-group if every primitive permutation group having a regular (transitive) subgroup isomorphic to B is $2$-transitive.

Schur proved that a cyclic group of composite order is a B-group. Wielandt showed that no group of the form $B_1 \times \cdots \times B_d$ with $|B_1| = \ldots = |B_d| \ge 3$ and $d \ge 2$ is a B-group. (See Theorems 25.3 and 25.7 in Wielandt, Finite permutation groups.)

Exercise 3.5.6 in Dixon and Mortimer Permutation groups asks for a proof that no elementary abelian $p$-group is a B-group. However this is false: the permutation groups of degree $4$ containing $C_2 \times C_2$ as a regular subgroup are $\langle (12)(34), (13)(24)\rangle$ and $\mathrm{Dih}(4)$ (both imprimitive) and $A_4$, $S_4$ (both $2$-transitive). My question is:

For which $d \in \mathbb{N}$ is the elementary abelian group $C_2^d$ a B-group?

Necessary conditions for a primitive permutation groups to have a regular subgroup are given in Corollary 3 of Liebeck, Praeger, Saxl, Transitive Subgroups of Primitive Permutation Groups, J. Alg 234, 291–361 (2000). The main theorem of Li, The finite primitive permutation groups containing an abelian regular subgroup, Proc. London Math. Soc. 87, 725–747 (2003), gives a sharper result. It seems non-trivial to use either of these papers to answer the question, but I'd welcome a correction.

If $H \le \mathrm{GL}(\mathbb{F}_2^d)$ then the subgroup $\mathbb{F}_2^d \rtimes H$ of the affine general linear group $\mathrm{AGL}(\mathbb{F}_2^d)$ is primitive if and only if $H$ is irreducible. So if there exists an irreducible $H$ acting intransitively on $\mathbb{F}_2^d \backslash \{0\}$, then $C_2^d$ is not a B-group. Such groups exist whenever $d$ is composite, but not always when $d$ is prime. For example, the three irreducible subgroups of $\mathrm{GL}(\mathbb{F}_2^3)$ all contain a Singer element of order $7$, so give $2$-transitive subgroups of $\mathrm{AGL}(\mathbb{F}_2^3)$. (The only other primitive permutation groups of degree $8$ containing a subgroup isomorphic to $C_2 \times C_2 \times C_2$ are $A_8$ and $S_8$, so it follows that $C_2 \times C_2 \times C_2$ is a B-group.) This motivates my second question:

For which primes $p$ is there an irreducible subgroup $H$ of $\mathrm{GL}(\mathbb{F}_2^p)$ such that $H$ acts intransitively on $\mathbb{F}_2^p \backslash \{0\}$?

  • 1
    $\begingroup$ In your second question, if $2^p-1$ is not prime, then there exists a cyclic group $H$ of order $k$ acting irreducibly, where $k$ is a proper of $2^p-1$ that does not divide $2^k-1$ for any $k<p$. So you can restrict attention to primes $p$ with $2^p-1$ prime.. I have checked using tables of known representations that there is no such $H$ for $p$ when $2^p-1$ is prime and $p \le 127$. Also, from the tables of primitive groups, $C_2^d$ is a $B$-group when $d=2,3,5,7,13$. But I don't know how to proceed with larger Mersenne primes. $\endgroup$ – Derek Holt Dec 30 '16 at 22:21
  • 2
    $\begingroup$ Re the second question C. Hering classified finite linear groups transitive on non-zero vectors- I find his papers difficult to read though. $\endgroup$ – Geoff Robinson Dec 31 '16 at 12:04
  • $\begingroup$ @GeoffRobinson The question here seems to be about the opposite problem - when are there intransitive irreducible subgroups? When $p$ and $2^p-1$ are both prime, such a subgroup of ${\rm GL}(p,2)$ would have to be in Aschbacher's class ${\mathcal S}$, i.e. an almost simple group. $\endgroup$ – Derek Holt Dec 31 '16 at 13:23
  • $\begingroup$ @DerekHolt : Well, yes, it is the opposite problem, but if you understand all the transitive (necessarily irreducible) linear groups, you understand all the intransitive irreducible linear groups as well. $\endgroup$ – Geoff Robinson Dec 31 '16 at 13:26
  • $\begingroup$ Mark, nice question. Just wondering about the provenance of the name "$B$-group" -- any ideas? $\endgroup$ – Nick Gill Jan 2 '17 at 15:32

This is a bit long for a comment. To answer the question it would be very useful to know what groups can act primitively on a set of order $2^a$. Looking at O'Nan--Scott--Aschbacher, we see that such groups must be either affine or almost simple, or product action. Let's consider these three families one at a time.

Almost simple. In this case, the following paper is useful:

Robert M. Guralnick, MR 700286 Subgroups of prime power index in a simple group, J. Algebra 81 (1983), no. 2, 304--311.

The main theorem of the paper classifies the objects in the title. A very useful corollary for us is this:

Corollary 2. Let $G$ be a nonabelian simple group acting transitively on $\Omega$ with $|\Omega|=p^a$ for some prime $p$. Then $G$ acts $2$-transitively on R unless $G \cong {\rm PSU}_4(2)$ and $|\Omega| = 27$ in which case $G$ acts as a rank $3$ primitive permutation group and $G$ has orbits of size $1$, $10$, and $16$.

This result means that, for the purposes of the question we can forget about simple groups containing regular elementary abelian subgroups -- they will never stop these subgroups being $B$-groups.

Affine. Now, as you point out, this reduces to studying irreducible subgroups $S$ of ${\rm GL}_a(2)$ that are intransitive on non-zero vectors. As Derek says we can assume that $a$ is prime and that $2^a-1$ is prime (otherwise $E_{2^a}$ cannot be a $B$-group).

Now, as Derek says, looking at Aschbacher's classification of subgroups of ${\rm GL}_a(q)$ (or maybe Nori's results would suffice), we can assume that $S$ is almost simple. So we're interested in almost simple groups $S$ that have a representation over $\mathbb{F}_2$ of degree a prime $p$ with $2^p-1$ prime, and are intransitive on non-zero vectors.

We can deal with that last condition -- intransitivity on non-zero vectors -- by referring to the main theorem in Guralnick's paper cited above. If $S$ is transitive on non-zero vectors, then this is a primitive action of prime degree and so is on Guralnick's list. I think with a little work you can probably rule out (nearly) all of the possibilities on this list, i.e. (with some possible exceptions) any almost simple group will be intransitive on vectors.

Product action. In this case the group $G\leq U\wr S_k$ where $U$ is almost simple with an action of degree $2^b$ with $b$ a proper divisor of $a$. In particular $a$ is composite, and we already know that groups $E_{2^a}$ with $a$ composite cannot be $B$-groups.

CONCLUSION. The only outstanding case is the affine one which means that we've reduced Question 1 to Question 2. And, given the progress on the affine case described above, based on Derek's remarks we would conclude the following (modulo the exceptions in the final paragraph of the affine case):

$E_{2^a}$ is a $B$-group if and only if $a$ is prime, $2^a-1$ is prime, and there exists no almost simple group (except ${\rm GL}_a(2)$) with an irreducible representation of degree $a$ over $\mathbb{F}_2$.

The problem of ascertaining when almost simple groups have such representations is currently beyond me.

  • $\begingroup$ I've realised that I need to deal with the product action as well. I think is pretty straightforward. Edit to follow... $\endgroup$ – Nick Gill Jan 4 '17 at 11:00
  • $\begingroup$ I guess Bob Guralnick would be the obvious person to ask about the outstanding case! $\endgroup$ – Derek Holt Jan 4 '17 at 11:22
  • $\begingroup$ Thank you very much. I'll have to think about this carefully, but it seems to go a long way to answering my question. $\endgroup$ – Mark Wildon Jan 4 '17 at 19:56

This post gives a complete proof of the final box in Nick Gill's excellent answer. Let $V = \mathbb{F}_2^n$.$\newcommand{\GL}{\mathrm{GL}}$

Theorem. $C_2^n$ is a B-group if and only if $2^n-1$ is a Mersenne prime and the only simple groups having an $n$-dimensional irreducible representation over $\mathbb{F}_2$ are $C_{2^n-1}$ and $\GL(V)$.

The key results we need are

  1. A transitive permutation group of prime degree is either solvable (and so affine) or $2$-transitive; this is due to Burnside.
  2. If $H \le \GL(V)$ is $2$-transitive then either $n=4$ and $H \cong A_7$ or $H = \GL(V)$; this was proved by Cameron and Kantor in 1979.
  3. An almost simple group of degree $2^n$ is $2$-transitive; this follows from Guralnick's Theorem in Nick's answer.

Proof. Suppose $2^n-1$ is composite. Let $g \in \mathrm{GL}(V)$ be a Singer element of order $2^n-1$. By Zsigmondy's Theorem, there is a power $g^r$ such that $\langle g^r \rangle$ acts irreducibly but intransitively on $V$. (This was noted by Derek Holt in a comment on the question.) Hence $V \rtimes \langle g^r \rangle$ is simply transitive and primitive and so $C_2^n$ is not a B-group.

Conversely, suppose $2^n-1$ is prime and that $G$ is a primitive permutation group of degree $2^n$. By the O'Nan–Scott theorem, $G$ is either almost simple, or of affine type, or of diagonal type, or of twisted wreath or product type. Following Nick's answer, in the final two cases, we have $G \le S_a \wr S_b$ with $a^b = 2^n$. If $a=2^m$ then $n= mb$. But $n$ is prime, so this is impossible. The diagonal case does not arise because $2^n$ is not the order of a simple group. By (3) if $G$ is almost simple then $G$ is $2$-transitive.

We are left with the affine case. Suppose that the simple group $S$ has an irreducible $n$-dimensional representation over $\mathbb{F}_2$. Then $V \rtimes S$ is primitive and simply transitive, so $C_2^n$ is not a B-group.

Finally, suppose that $C_2^n$ is not a B-group. We must find a proper non-cyclic irreducible simple subgroup $S$ of $\GL(V)$. We are in the affine case, so there exists an irreducible $H \le \GL(V)$ such that $H$ is intransitive on $V \backslash \{0\}$. Let $M$ be a maximal subgroup of $\GL(V)$ containing $H$. The 10 irreducible Aschbacher classes all involve groups preserving a decomposition of $V$ that can exist only when $\dim V = n$ is composite. Therefore $M$ is almost simple.

If $M$ is transitive on $V \backslash \{0\}$ then, since $|V \backslash \{0\}| = 2^n-1$ is prime, (1) implies that either $M \cong C_{2^n-1} \rtimes C_s$ for some $s$ dividing $2^n-1$, or $M$ is $2$-transitive. In the first case, since $H$ is irreducible, $H \ge C_{2^n-1}$. In the remaining case $V \rtimes M$ is $3$-transitive, and by (2), $M = \GL(V)$, and so $H = \GL(V)$. In both cases this contradicts the intransitivity of $H$. Hence we did not lose simple transitivity in passing from $H$ to $M$. Let $S$ be the simple subgroup of $M$. By Clifford's Theorem, the restriction of the $\mathbb{F}_2 M$-module $V$ to $S$ is a direct sum of conjugate irreducible submodules. But $\dim V = n$ is prime. Hence $S$ acts irreducibly. Since $M$ is intransitive on $V \backslash \{0\}$, so is $S$. Hence $S$ is as required. $\Box$

Remark. It seems plausible that the only simple groups having an $n$-dimensional irreducible representation over $\mathbb{F}_2$ for $n$ such that $2^n-1$ is a Mersenne prime are $C_{2^n-1}$ and $\GL(V)$. From the modular Atlas data in GAP, the only sporadic simple groups that could be exceptions are $J_4$, $Ly$, $Th$, $Fi_{24}$, $B$ and $M$. Tables of small dimensional representations of quasi-simple groups in non-defining characteristic and Chevalley groups in defining characteristic, show there are no exceptional representations if $n \le 250$. So if $n \le 250$ then $C_2^n$ is a B-group if and only if

$$ n \in \{1,2,3,5,7,13,17,19,31,61,89,107,127\}. $$

Edit. I had to edit this answer to remove the entirely false claim in the final remark that all irreducible representations of alternating groups over $\mathbb{F}_2$ are self-dual (and so have an invariant symplectic form and are even dimensional). The isomorphism $A_8 \cong \mathrm{GL}_4(\mathbb{F}_2)$ shows this is false.

  • $\begingroup$ I don't understand your comment at the end "barring a remarkable low-dimensional representation ...". According to the Hiss-Malle tables there is no such representation. $\endgroup$ – Derek Holt Apr 26 '17 at 13:21
  • $\begingroup$ You are quite right. Also I forgot about the alternating groups. I'll edit the answer. $\endgroup$ – Mark Wildon Apr 26 '17 at 14:09
  • $\begingroup$ Mark, I just saw this answer. Good stuff! The representation theory question that is left looks rather intriguing... I only wish I could say something useful about it... $\endgroup$ – Nick Gill May 3 '17 at 7:58

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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