Regular abelian subgroups of permutation groups

A B-group is a group K such that if G is a permutation group containing K as a regular subgroup then G is either imprimitive or 2-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 p is an odd prime then C_{p^2} 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 2-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 G is imprimitive, it does so by showing that the kernel N of an irreducible non-trivial constituent of the permutation character of G is intransitive. The orbits of N are then a non-trivial block system for G. Thus G 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 G is a permutation group containing a regular abelian subgroup K then G is primitive if and only if G is quasiprimitive.

Proof. Suppose B is part of a non-trivial block system \mathcal{B} for G. Let L be the kernel of the induced action of K on \mathcal{B}. The key observation is that K/L acts regularly on \mathcal{B}: it acts semiregularly, since K/L is abelian, and since K is transitive, K/L is transitive on \mathcal{B}. Similarly, since K is transitive, L is transitive on the block B (given \alpha, \beta \in B, there exists k \in K such that \alpha^k = \beta; then B^k = B and because k acts regularly on \mathcal{B}, we have k \in L). So we can factor K as a top part K/L, acting regularly on \mathcal{B}, and a bottom part L acting regularly on each block. In particular, the kernel of G acting on \mathcal{B} contains L and L has \mathcal{B} as its orbits. Therefore \mathcal{B} is the set of orbits of a normal subgroup of G, and G is not quasiprimitive. \Box

Dropping the condition that K be abelian, one finds many imprimitive but quasiprimitive groups: they can be obtained from factorizations G = HK of a finite simple group G where H \cap K = 1 and H is not maximal. One nice example is A_7 = F_{21}S_5 where F_{21} is the transitive Frobenius group of order 21 inside A_7 and S_5 is generated by (1,2,3,4,5) and (1,2)(6,7). Here F_{21} is not maximal because it is a subgroup of \mathrm{GL}_3(\mathbb{F}_2) in its action on the 7 non-zero vectors of \mathbb{F}_2^3. For an easier example, take A_5 = \langle (1,2,3,4,5) \rangle A_4.

Factorizable groups

Say that a finite group K is factorizable if there exists t > 1 and groups K_1, \ldots, K_t such that K \cong K_1 \times \cdots \times K_t where |K_1| = \ldots = |K_t| \ge 3. If K is factorizable with |K_i| = m for each i, then, since each K_i acts as a regular subgroup of S_m, K is a regular subgroup of S_m \wr S_t in its primitive action on \{1,\ldots,m\}^t. 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 G = V \rtimes H where V = \mathbb{F}_p^n for some prime p and H \le \mathrm{GL}(V). Here V acts by translation on itself and H is the point stabiliser of 0. Any non-trivial block is invariant under the action of V and so is a linear subspace of V. Therefore G is primitive if and only if H acts irreducibly on V. The action of G is 2-transitive if and only if the point stabiliser H is transitive. Therefore if there exists a linear group H acting irreducibly but not transitively on \mathbb{F}_p^n, then C_p^n is not a B-group.

Symmetric group representations

Consider the symmetric group S_n acting on \langle v_1, \ldots, v_n \rangle by permutation matrices. Let V = \langle v_i - v_j : 1 \le i  2 and n \ge 2 then the orbit of v_1-v_2 does not contain v_2-v_1, and hence C_p^n is not a B-group in these cases. If p = 2 and n \ge 4 then the orbit of v_1+v_2 does not contain v_1+v_2+v_3+v_4. Therefore C_2^m is not a B-group when m is even and m \ge 4.

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 V \rtimes H may have regular abelian subgroups other than the obvious translation subgroup V. Let t_v : V \rightarrow V denote translation by v \in V.
Take n=2r+1, fix s \le r, and consider the subgroup

K = \langle (2,3)t_{v_1+v_2}, \ldots, (2s,2s+1)t_{v_1+v_{2s}}, t_{v_{2s+2}}, \ldots, t_{v_{2r+1}} \rangle .

By the multiplication rule

h t_v h' t_{v'} = h h' t_{vh' + v'}

we see that (2t,2t+1)t_{v_1+v_{2t}} has square t_{v_{2t}+v_{2t+1}} and that the generators of K commute. Therefore K \cong C_4^s \times C_2^{2(r-s)}, and no group of this form, with r \ge 2 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 n is odd, needed to ensure that V is irreducible, is omitted. In fact it is an open problem to decide when C_2^n 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 K is a regular abelian subgroup of G then for each v \in V, there exists a unique g_v \in K such that 0g_v = v. There exists a unique h_v \in H such that g_v = h_v t_v. By the multiplication rule above we have

h_u t_u h_v t_v = h_u h_v t_{uh_v + v} = h_vh_u t_{vh_u+u} = h_vt_vh_ut_u

for all u, v \in V. Therefore \{h_v : v \in V\} is an abelian subgroup of H and uh_v + v = vh_u + u for all u,v \in V. Replacing v with v+w, we have

\begin{aligned} uh_{v+w} + (v+w) &= (v+w)h_u + u \\                  &= vh_u + wh_u + u \\                  &= uh_v + v + uh_w + w - u \end{aligned}

and so, cancelling v+w, we get the striking linearity property

h_{v+w} = h_v + h_w - \mathrm{id}.

Equivalently, h_{v+w}-\mathrm{id} = (h_v-\mathrm{id}) + (h_w-\mathrm{id}). Since (h_v - \mathrm{id})(h_w - \mathrm{id}) = (h_vh_w - \mathrm{id}) - (h_v - \mathrm{id}) - (h_w - \mathrm{id}), it follows that the linear maps \{h_v - \mathrm{id} : v \in V \} form a subalgebra of \mathrm{End}(V). (This is essentially Fact 3 in the linked paper.)

Abelian regular subgroups of odd degree affine symmetric groups

Suppose that K is a regular abelian subgroup of V \rtimes S_n and that there exists v \in V such that h_v \not= \mathrm{id}. The matrix M representing h_v in the basis v_2-v_1, \ldots, v_n-v_1 of V is a permutation matrix if 1 h_v = 1. If 1h_v = a > 1 and bh_v = 1 then from

(v_i - v_1)h_v = (v_{ih}-1) - (v_a-v_1)

we see that M has -1 in each entry in the column for v_a-v_1, and the only other non-zero entries are a unique 1 in the row for v_i-v_1, for each i \not= b. For example, (1,2,3,4) is represented by

\left( \begin{matrix} -1 & 1 & 0 \\ -1 & 0 & 1 \\ -1 & 0 & 0 \end{matrix} \right).

By the linearity property above, h_{2v} = 2h_v - \mathrm{id}. But if p > 2 then 2M - I is not of either of these forms. Therefore V is the unique regular abelian subgroup of V \rtimes S_n.

Suppose that p = 2. Suppose that h_v has a cycle of length at least 4. By relabelling, we may assume this cycle is (1,2,\ldots,a) where a is a 2-power. The matrix representing h_v has -1 entries in column 2, and the matrix representing h_v^{-1} has -1 entries in column a. Therefore the matrix representing h_v + h_v^{-1} + \mathrm{id} has -1 entries in both columns 2 and a, and so is not of the permitted form. It follows that, when p=2 and n=2r+1 is odd, the only abelian regular subgroups of V \rtimes S_n are isomorphic to C_4^s \times C_2^{2r-2s}. (In fact it seems they are precisely the subgroups constructed above.)

Abelian regular subgroups of even degree affine symmetric groups

Now suppose that n is even. Let U = V / \langle v_1 + \cdots + v_n \rangle and consider U \rtimes S_n. When n=4 the action of S_4 is not faithful: after factoring out the kernel \langle (1,2)(3,4), (1,3)(2,4)\rangle we get \mathbb{F}_2^2 \rtimes S_3 \cong S_4, which has abelian regular subgroups C_2 \times C_2 and C_4. When n=6, there is an abelian regular subgroup isomorphic to C_8 \times C_2, generated by t_{v_1+v_3}(3,4,5,6) and t_{v_1+v_2}. The obstruction seen above does not apply: for example, in the basis v_1+v_3, v_1+v_4,v_1+v_5,v_1+v_6, we have

(3,4,5,6) + (3,6,5,4) + \mathrm{id} \mapsto \left( \begin{matrix} 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \end{matrix} \right)

which is the matrix representing (1,2)(3,6,5,4), thanks to the relations present in the quotient module. However, in this case the action of S_6 on \mathbb{F}_2^4 is transitive (any element of U is congruent to some v_i + v_j) and in fact C_8 \times C_2 is a B-group. (A related, remarkable fact, is that the action of A_6 on U extends to a 2-transitive action of A_7, giving an example of a 3-transitive affine group, U \rtimes A_7.) A computer calculation shows that, up to conjugacy, there are also three abelian regular subgroups of U \rtimes S_n isomorphic to C_4 \times C_4 and two isomorphic to C_4 \times C_2 \times C_2. If n \ge 8 then a similar argument to the odd degree case shows that any abelian regular subgroup has exponent at most 4, so we get no new examples.

Elementary abelian subgroups other than V

The following lemma has a simple ad-hoc proof.

Lemma. If n < p then the only regular abelian subgroups of V \rtimes \mathrm{GL}(V) are elementary abelian.

Proof. Suppose that K is an elementary abelian subgroup of V \rtimes \mathrm{GL}(V). Let h_v t_v \in K. Since h_v - 1 is a nilpotent n \times n-matrix and n \textless\; n we have (h_v - 1)^p = 0 The p-th power of h_v t_v \in V is therefore t_w where w = v + vh_v + \cdots + vh_v^{p-1}. Since each Jordan block in h_v has size at most p-1, we have w=0. Therefore K has exponent p. \Box

Note however that there exist elementary abelian subgroups other than the obvious V whenever n \ge 2. Possibly they can be classified by the correspondence with nil-algebras in the Caranti, Della Volta, Sala paper.

Algebra groups

Given n \in \mathbb{N} and a field F, define an algebra group to be a subgroup of \mathrm{GL}_n(F) of the form \{I + X : X \in A \} where A is a nilalgebra of n \times n matrices. (We may assume all elements of A are strictly upper triangular.) The subalgebra property above shows that any abelian regular subgroup of an affine group of dimension n over \mathbb{F}_2 is an extension

1 \rightarrow C_2^r \hookrightarrow K \twoheadrightarrow \{I + X : X \in A \} \rightarrow 1

of an algebra group by an elementary abelian 2-group. If F has characteristic 2 then, since

(I+X)(I+Y) = (I+XY) + (I+X) + (I+Y)

by the dual of the calculation above, a subgroup G of \mathrm{GL}_n(F) is an algebra group if and only if \{ I + g : g \in G \} is additively closed.

An exhaustive search shows that if n \le 4 then every 2-subgroup of \mathrm{GL}_n(\mathbb{F}_2) is an algebra group; note it suffices to consider subgroups up to conjugacy. But when n = 5 there are, up to conjugacy, 110 algebra subgroups of \mathrm{GL}_5(\mathbb{F}_2), and 66 non-algebra subgroups, of which 34 are not isomorphic to an algebra subgroup. Of these:

  • 1 is abelian, namely C_8;
  • 4 are nilpotent of class 2, namely \langle x, c : x^8 = c^2 = 1, x^c = x^5 \rangle, \langle x, y, c : x^4 = y^2 = c^4 = 1, x^c = xy, y^c = y \rangle, \langle x, y, c : x^2 = y^2 = c^8 = 1, x^c = xy, y^c = y \rangle, \langle x,y,c : x^4 = y^2 = c^4 = 1, x^c = xy, y^c = y\rangle;
  • 17 are nilpotent of class 3;
  • 12 are nilpotent of class 4.

The appearance of C_8 is explained by the following lemma, which implies that there is an algebra group isomorphic to C_{2^m} if and only if m \le 2.

Lemma. Let g \in \mathrm{GL}_n(\mathbb{F}_2) have order 2^m. Then \langle g \rangle is an algebra group if and only if all Jordan blocks in g have dimension at most 3.

Proof. Let 1 + X be a Jordan block of g of dimension d and order 2^m. The subalgebra of d\times d-matrices generated by X has dimension d-1, and so has size 2^{d-1}. Therefore if \langle g \rangle is an algebra group then 2^{d-1} \le 2^m. Hence d-1 \le m. On the other hand, since X^{d-1} \not=0, if 2^r < d then (1+X)^{2^r} \not=0, and so the order of 1+X is the minimal s such that 2^s \ge d. Hence 2^{m-1} < d. Combining these inequalities we get

2^{m-1} < d \le m+1

and so m \le 2, and, since d-1 \le m, we have d \le 3. The converse is easily checked. \Box.


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