Regular abelian subgroups of permutation groups
Wildon's Weblog 2018-03-12
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.
Quasiprimitivity
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
.
Factorizable groups
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.
Affine 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.
Algebra groups
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.
.