Regular abelian subgroups of permutation groups

Wildon's Weblog 2018-03-12

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.

Quasiprimitivity

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. $\Box$ .