A B-group is a group such that if is a permutation group containing as a regular subgroup then is either imprimitive or -transitive. (Regular subgroups are always transitive in this post.) The term ‘B-group’ was introduced by Wielandt, in honour of Burnside, who showed in 1901 that if is an odd prime then is a B-group. This is a companion to his important 1906 theorem that a transitive permutation group of prime degree is either solvable or -transitive.
One might imagine that, post-classification, it would be clear which groups are B-groups, but this is far from the case. The purpose of this post is to collect some ancillary results with a bearing on this question.
When Burnside’s argument succeeds in proving that is imprimitive, it does so by showing that the kernel of an irreducible non-trivial constituent of the permutation character of is intransitive. The orbits of are then a non-trivial block system for . Thus is not even quasiprimitive. That one always gets this stronger result is explained by Corollary 3.4 in this paper by Li. The details are spelled out below.
Claim. If is a permutation group containing a regular abelian subgroup then is primitive if and only if is quasiprimitive.
Proof. Suppose is part of a non-trivial block system for . Let be the kernel of the induced action of on . The key observation is that acts regularly on : it acts semiregularly, since is abelian, and since is transitive, is transitive on . Similarly, since is transitive, is transitive on the block (given , there exists such that ; then and because acts regularly on , we have ). So we can factor as a top part , acting regularly on , and a bottom part acting regularly on each block. In particular, the kernel of acting on contains and has as its orbits. Therefore is the set of orbits of a normal subgroup of , and is not quasiprimitive.
Dropping the condition that be abelian, one finds many imprimitive but quasiprimitive groups: they can be obtained from factorizations of a finite simple group where and is not maximal. One nice example is where is the transitive Frobenius group of order inside and is generated by and . Here is not maximal because it is a subgroup of in its action on the non-zero vectors of . For an easier example, take .
Say that a finite group is factorizable if there exists and groups such that where . If is factorizable with for each , then, since each acts as a regular subgroup of , is a regular subgroup of in its primitive action on . Therefore no factorizable group is a B-group. This example appears as Theorem 25.7 in Wielandt Permutation Groups.
An interesting source of examples is affine groups of the form where for some prime and . Here acts by translation on itself and is the point stabiliser of . Any non-trivial block is invariant under the action of and so is a linear subspace of . Therefore is primitive if and only if acts irreducibly on . The action of is 2-transitive if and only if the point stabiliser is transitive. Therefore if there exists a linear group acting irreducibly but not transitively on , then is not a B-group.
Symmetric group representations
Consider the symmetric group acting on by permutation matrices. Let and then the orbit of does not contain , and hence is not a B-group in these cases. If and then the orbit of does not contain . Therefore is not a B-group when is even and .
Exotic regular abelian subgroups of affine groups
An interesting feature of these examples, noted by Li in Remark 1.1 following the main theorem, is that may have regular abelian subgroups other than the obvious translation subgroup . Let denote translation by .
Take , fix , and consider the subgroup
By the multiplication rule
we see that has square and that the generators of commute. Therefore , and no group of this form, with is a B-group. (Since these groups are factorizable, this also follows from Wielandt’s result above.)
Li’s paper includes this example: the assumption that is odd, needed to ensure that is irreducible, is omitted. In fact it is an open problem to decide when is a B-group.
Perhaps surprisingly, this example does not generalize to odd primes. To see this, we introduce some ideas from an interesting paper by Caranti, Della Volta and Sala. Observe that if is a regular abelian subgroup of then for each , there exists a unique such that . There exists a unique such that . By the multiplication rule above we have
for all . Therefore is an abelian subgroup of and for all . Replacing with , we have
and so, cancelling , we get the striking linearity property
Equivalently, . Since , it follows that the linear maps form a subalgebra of . (This is essentially Fact 3 in the linked paper.)
Abelian regular subgroups of odd degree affine symmetric groups
Suppose that is a regular abelian subgroup of and that there exists such that . The matrix representing in the basis of is a permutation matrix if . If and then from
we see that has in each entry in the column for , and the only other non-zero entries are a unique in the row for , for each . For example, is represented by
By the linearity property above, . But if then is not of either of these forms. Therefore is the unique regular abelian subgroup of .
Suppose that . Suppose that has a cycle of length at least . By relabelling, we may assume this cycle is where is a 2-power. The matrix representing has entries in column , and the matrix representing has entries in column . Therefore the matrix representing has entries in both columns and , and so is not of the permitted form. It follows that, when and is odd, the only abelian regular subgroups of are isomorphic to . (In fact it seems they are precisely the subgroups constructed above.)
Abelian regular subgroups of even degree affine symmetric groups
Now suppose that is even. Let and consider . When the action of is not faithful: after factoring out the kernel we get , which has abelian regular subgroups and . When , there is an abelian regular subgroup isomorphic to , generated by and . The obstruction seen above does not apply: for example, in the basis , we have
which is the matrix representing , thanks to the relations present in the quotient module. However, in this case the action of on is transitive (any element of is congruent to some ) and in fact is a B-group. (A related, remarkable fact, is that the action of on extends to a -transitive action of , giving an example of a -transitive affine group, .) A computer calculation shows that, up to conjugacy, there are also three abelian regular subgroups of isomorphic to and two isomorphic to . If then a similar argument to the odd degree case shows that any abelian regular subgroup has exponent at most , so we get no new examples.
Elementary abelian subgroups other than
The following lemma has a simple ad-hoc proof.
Lemma. If then the only regular abelian subgroups of are elementary abelian.
Proof. Suppose that is an elementary abelian subgroup of . Let . Since is a nilpotent -matrix and we have The -th power of is therefore where . Since each Jordan block in has size at most , we have . Therefore has exponent .
Note however that there exist elementary abelian subgroups other than the obvious whenever . Possibly they can be classified by the correspondence with nil-algebras in the Caranti, Della Volta, Sala paper.
Given and a field , define an algebra group to be a subgroup of of the form where is a nilalgebra of matrices. (We may assume all elements of are strictly upper triangular.) The subalgebra property above shows that any abelian regular subgroup of an affine group of dimension over is an extension
of an algebra group by an elementary abelian 2-group. If has characteristic then, since
by the dual of the calculation above, a subgroup of is an algebra group if and only if is additively closed.
An exhaustive search shows that if then every 2-subgroup of is an algebra group; note it suffices to consider subgroups up to conjugacy. But when there are, up to conjugacy, 110 algebra subgroups of , and 66 non-algebra subgroups, of which 34 are not isomorphic to an algebra subgroup. Of these:
- 1 is abelian, namely ;
- 4 are nilpotent of class 2, namely , , , ;
- 17 are nilpotent of class 3;
- 12 are nilpotent of class 4.
The appearance of is explained by the following lemma, which implies that there is an algebra group isomorphic to if and only if .
Lemma. Let have order . Then is an algebra group if and only if all Jordan blocks in have dimension at most .
Proof. Let be a Jordan block of of dimension and order . The subalgebra of -matrices generated by has dimension , and so has size . Therefore if is an algebra group then . Hence . On the other hand, since , if then , and so the order of is the minimal such that . Hence . Combining these inequalities we get
and so , and, since , we have . The converse is easily checked. .