Burnside’s method

April 24, 2017

Burnside proved in 1901 that if p is an odd prime then a permutation group containing a regular subgroup isomorphic to C_{p^2} is either imprimitive or 2-transitive. His proof was an early application of character theory to permutation groups. Groups with this property are now called B-groups.

Burnside attempted to generalize his 1901 result in two later papers: in 1911, he claimed a proof that C_{p^n} is a B-group for any prime p and any n \ge 2, and in 1921, he claimed a proof that all abelian groups, except for elementary abelian groups, are B-groups. The first claim is correct, but his proof has a serious gap. This error appears to have been unobserved (or, just possibly, observed but ignored, since the result was soon proved in another way using Schur’s theory of S-rings) until 1994 when it was noted by Peter Neumann, whose explication may be found in his introduction to Burnside’s collected works. In 1995, Knapp extended Burnside’s argument to give a correct proof. Burnside’s second claim is simply false: for example, S_4 \wr S_2 acts primitively on \{1,2,3,4\}^2, and has a regular subgroup isomorphic to C_4 \times C_4. In one of my current projects, I’ve simplified Knapp’s proof and adapted Burnside’s character-theoretic methods to show, more generally, that any cyclic group of composite order is a B-group.

The purpose of this post is to record some proofs omitted for reasons of space from the draft paper. This companion post has some notes on B-groups that may be of more general interest.

Sums over roots of unity

Let \xi be a primitive nth root of unity. Define

R(r) = \{ r, r + p^{n-1}, \ldots, r+ (p-1)p^{n-1} \}

for 0 < r < p^{n-1}. Define a subset Z of \{1,\ldots,p^n-1\} to be null if there exists s \in \mathbb{N}_0 and distinct r_{ij} \in \{1,\ldots, p^{n-1}-1\} for 0 \le i \le p-1 and 1 \le j \le s such that r_{ij} \equiv i mod p for each i and j and

Z = \bigcup_{i=0}^{p-1} \bigcup_{j=1}^s R(r_{ij}).

Proposition 6.2 Let \omega = \zeta^{p^{n-1} c} where c is not divisible by p. Let \mathcal{O} \subseteq \{1,\ldots,p^n-1\}. Then

\sum_{i \in \mathcal{O}} \zeta^i = \sum_{i \in \mathcal{O}} \omega^i

if and only if either

  1. \mathcal{O} is null; or
  2. \mathcal{O} = \{p^{n-1}, \ldots, (p-1)p^{n-1} \} \; \cup \; \bigcup_{i=1}^{p-1} R(r_i) \; \cup \; Z where Z is a null set, the r_i are distinct elements of \{1,\ldots,p^{n-1}-1\}\backslash Z and r_i \equiv i mod p for each i.

Proof. Since the minimum polynomial of \zeta is

1+X^{p^{n-1}} + \cdots + X^{(p-1)p^{n-1}}

we have \sum_{i \in R(r)} \zeta^i = 0. Since \xi^{p^{n-1}} = \omega, we have \sum_{i \in R(r)} \omega^i = p\omega^r. It follows that \sum_{i \in Z} \xi^i = \sum_{i \in Z} \omega^i = 0 if Z is a null set. (For the second equality, note the contributions from the R(r_{ij}) for fixed j combine to give p + p\omega + \cdots + p\omega^{p-1} = 0.) For (2) we have \sum_{i=1}^{p-1} \xi^{i p^{n-1}} = \omega + \cdots + \omega^{p-1} = -1, \sum_{i=1}^{p-1} \omega^{i p^{n-1}} = (p-1) and \sum_{i=1}^{p-1} p \omega^i = - p. This proves the ‘if’ direction.

Conversely, by Lemma 2.1 in the paper, \mathcal{O} \backslash \{p^{n-1},\ldots,(p-1)p^{n-1} \} is a union of some of some of the sets R(r). There exists a unique subset A of \{1,\ldots,p-1\} and unique j_i \in \mathbb{N} for 0 \le i \le p-1 and unique r_{ij} \in \{1,\ldots, p^{n-1}-1\} for 1 \le j \le s and 0 \le i \le j_i such that

\mathcal{O} = \{ p^{n-1} i : i \in A \} \cup \bigcup_{i=0}^{p-1}\bigcup_{j=1}^{i_j} R(r_{ij}).

We have \sum_{i \in \mathcal{O}} \zeta^i = \sum_{i \in A} \omega^i and \sum_{i \in \mathcal{O}} \omega^i = |A| + \sum_{i=0}^{p-1} pj_i \omega^i. Therefore

|A| + \sum_{i=0}^{p-1} (pj_i - [i \in A]) X^i

has \omega as a root. Since this polynomial has degree at most p-1 and the minimal polynomial of \omega is 1+X +\cdots + X^{p-1}, it follows that the coefficients are constant. Hence

|A| + pj_0 = pj_i - [i \in A]

for 1\le i \le p-1. If A = \varnothing then j_0 = j_1 = \ldots = j_{p-1} and \mathcal{O} is null. Otherwise, taking the previous displayed equation mod p we see that A = \{1,\ldots,p-1\}, and, moreover, the j_i are constant for 1 \le i \le p-1. (This holds even if p=2.) Set

s = j_1 = \ldots = j_{p-1}.

We have j_0 = s-1. Hence, s \in \mathbb{N} and choosing in any way r_i \in \{1,\ldots,p^{n-1}-1\} for 1 \le i \le p-1 such that r_i \in \mathcal{O}, we see that \mathcal{O} is the union of \{p^{n-1}, \ldots, (p-1)p^{n-1}\}, the sets R(r_i) for 1 \le i \le p-1 and a null set. \Box.

Ramanujan matrices.

For p prime and n \in \mathbb{N} we define R(p^n) to be the matrix

\left( \begin{matrix} \scriptstyle 1 &  \scriptstyle1 & \scriptstyle 1  &  \scriptstyle\ldots & \scriptstyle 1 &  \scriptstyle1 &  \scriptstyle1 \\  \scriptstyle-1 & \scriptstyle p-1 & \scriptstyle p-1  & \scriptstyle \ldots & \scriptstyle p-1 & \scriptstyle p-1 & \scriptstyle p-1 \\  \scriptstyle0 & \scriptstyle -p & \scriptstyle p(p-1)  & \scriptstyle \ldots &  \scriptstyle p(p-1) & \scriptstyle p(p-1) & \scriptstyle p(p-1) \\  \scriptstyle 0 & \scriptstyle 0 & \scriptstyle -p^2  & \scriptstyle \ldots & \scriptstyle p^2(p-1) & \scriptstyle p^2(p-1) & \scriptstyle p^2(p-1) \\  \scriptstyle\vdots & \scriptstyle \vdots & \scriptstyle \vdots  & \scriptstyle \ddots & \scriptstyle \vdots & \scriptstyle \vdots & \scriptstyle \vdots \\  \scriptstyle 0 & \scriptstyle 0 & \scriptstyle 0 & \scriptstyle \ldots & \scriptstyle -p^{n-2} & \scriptstyle p^{n-2}(p-1) & \scriptstyle  p^{n-2}(p-1) \\  \scriptstyle 0 & \scriptstyle 0 & \scriptstyle 0 & \scriptstyle \ldots & \scriptstyle 0 & \scriptstyle -p^{n-1} & \scriptstyle p^{n-1}(p-1) \end{matrix} \right).

More generally, if d has prime factorization p_1^{a_1} \ldots p_s^{a_s}, we define R(d) = R(p_1^{a_1}) \otimes \cdots \otimes R(p_s^{a_s}). The rows and columns of R(d) are labelled by the divisors D of 2^m p^n, as indicated below for the case d = 2^3 p, with p and odd prime.

R(2^3 p) = \begin{matrix} 1 \\ 2 \\ 2^2 \\ 2^3 \\ p \\ 2p \\ 2^2p \\ 2^3p \end{matrix} \left( \begin{array}{cccc|cccc}  1 & 1 & 1 & 1  & 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & 1 & -1 & 1 & 1 & 1 \\ 0  & -2  & 2 &2 & 0 & -2 & 2  & 2 \\ 0 & 0 & -2^2 & 2 & 0 & 0 & -2^2 & 2^2\\ \hline -1 & -1 & -1 & -1  & 1 & 1 & 1 & 1 \\ 1 & -1 & -1 & -1 & -1 & 1 & 1 & 1 \\ 0  & 2  & -2 & -2 & 0 & -2 & 2 & 2 \\ 0 & 0 & 2^2 & -2^2 & 0 & 0 & -2^2 & 2^2 \end{array} \right).

Say that a partition of D \backslash \{ d\} is coprime if the highest common factor of the numbers in each part is 1. The aim of the game is to find a non-empty set of rows, X say, of R(d) such that the subsets of columns (excluding column d) on which the sum of the rows in X are equal form a coprime partition of the divisors.

There is an application to B-groups only when d is even, in which case we may assume that 2 \in X. Proposition 6.7 in the paper states that in this case, the only way to win this game when d = 2^n, d = 2^n p or d = 2 p^n, where p is an odd prime, is to put every single row in X. This implies that C_{2^m p^n} is a B-group when n \le 1 or m \le 1. However the result on the game may well hold more generally.

Constant sums for p^n

The only coprime partition of D \backslash \{p^n\} has a singleton part, so the row sums are all equal. By adding p^{e-1} to each entry in row p^e for e \in \{1,\ldots, n\}, we obtain the matrix below.

\left( \begin{matrix} \scriptstyle 1 &  \scriptstyle1 & \scriptstyle 1  &  \scriptstyle\ldots & \scriptstyle 1 &  \scriptstyle1 &  \scriptstyle1 \\  \scriptstyle 0 & \scriptstyle p & \scriptstyle p  & \scriptstyle \ldots & \scriptstyle p & \scriptstyle p & \scriptstyle p \\  \scriptstyle p & \scriptstyle 0 & \scriptstyle p^2  & \scriptstyle \ldots &  \scriptstyle p^2 & \scriptstyle p^2 & \scriptstyle p^2  \\  \scriptstyle p^2 & \scriptstyle p^2 & \scriptstyle 0  & \scriptstyle \ldots & \scriptstyle p^3  & \scriptstyle p^3  & \scriptstyle p^3 \\  \scriptstyle\vdots & \scriptstyle \vdots & \scriptstyle \vdots  & \scriptstyle \ddots & \scriptstyle \vdots & \scriptstyle \vdots & \scriptstyle \vdots \\  \scriptstyle p^{n-2} & \scriptstyle p^{n-2} & \scriptstyle p^{n-2} & \scriptstyle \ldots & \scriptstyle 0 & \scriptstyle p^{n-1} & \scriptstyle  p^{n-1} \\  \scriptstyle p^{n-1} & \scriptstyle p^{n-1} & \scriptstyle p^{n-1} & \scriptstyle \ldots & \scriptstyle p^{n-1} & \scriptstyle 0 & \scriptstyle p^n \end{matrix} \right)

which still has constant sums over the rows in X. Let x_i = 1 if p^i \in X and let x_i = 0 otherwise. Writing numbers in base p, the row sums over X are, for each column,

\begin{aligned} 1 \quad &\quad x_nx_{n-1} \ldots x_3x_2x_0 \\ p \quad &\quad  x_n x_{n-1} \ldots x_3x_1x_0 \\ p^2\quad &\quad  x_n x_{n-1} \ldots x_2x_1x_0 \\ \vdots \; \quad &\quad  \qquad \vdots \\ p^{n-2} \quad &\quad  x_nx_{n-2} \ldots x_2x_1x_0 \\ p^{n-1} \quad &\quad  x_{n-1}x_{n-2} \ldots x_2x_1x_2. \end{aligned}

(Note in each case there are n digits, the most significant corresponding to p^{n-1}.) The numbers in the right column are equal, hence all the x_i are equal and X = D, as required. Taking p=2 this gives an alternative proof of Proposition 6.7(i).

Proofs of further claims on the Ramanujan matrices

Let R(d)^{\circ} denote R(d) rotated by a half turn. The following result was stated without proof in the paper; it implies that each R(d) is invertible, with inverse \frac{1}{d}R(d)^\circ. (Update: I later found a much better proof, now outlined in the paper, using a lower-upper decomposition of R(d).)

Proposition. We have R(p^n)^{-1} = \frac{1}{p^n} R(p^n)^\circ and \det R(p^n) = p^{n(n+1)}/2.

Proof. Since R(p^n)^\circ_{p^sp^t} = R(p^n)_{p^{n-s}p^{n-t}}, we have R(p^n)^\circ_{p^np^t} = 1 for all t \in \{0,\ldots, n\} and

\begin{aligned} R(p^n)^\circ_{p^sp^t} &= \begin{cases} (p-1)p^{n-s-1} & n-t \ge n-s  \\ -p^{n-s-1} & n-t=n-s-1 \\ 0 & n-t \le n-s-2 \end{cases} \\ &= \begin{cases} (p-1)p^{n-s-1} & s \ge t \\ -p^{n-s-1} & s = t-1 \\ 0 & s \le t-2 \end{cases} \end{aligned}

for s \in \{0,\ldots, n-1\} and t \in \{0,\ldots,n\}. We use this to show that

\sum_{c=0}^n R(p^n)_{p^rp^c} R(p^n)^{\circ}_{p^c p^{r'}} = p^n [r=r']

for r,r' \in \{0,\ldots, n\}. When r=0 the first term in each product is 1, so the left-hand side is the column sum of the n-r'-th of R(p^n); this is p^n if r'=0 and 0 otherwise, as required. Now suppose that r \ge 1. Since R(p^n)_{p^rp^c} vanishes when c \le r-2, the left-hand side is

-p^{r-1}R(p^n)^\circ_{p^{r-1}p^{r'}} + \sum_{c=r}^n p^{r-1}(p-1) R(p^n)_{p^cp^{r'}}.

Take out a factor p^{r-1} to define L. Substitute for R(p^n)^\circ_{p^cp^{r'}}, and split off the summand from R(p^n)^\circ_{p^np^{r'}} = 1 to get

\begin{aligned} L &= \begin{cases} (p-1)p^{n-r} & r' \le r-1 \\ -p^{n-r} & r' = r \\ 0 & r' \ge r+1 \end{cases} \\ & \qquad + (p-1) \sum_{c=r}^{n-1} (p-1) \begin{cases} (p-1)p^{n-c-1} & c \ge r' \\ -p^{n-c-1} & c = r'-1 \\ 0 & c \le r'-2 \end{cases} + p-1.\end{aligned}

We must now consider three cases. When r=r' we get

\begin{aligned} L &= p^{n-r} + (p-1)\sum_{c=r}^{n-1} (p-1)p^{n-c-1} + p-1 \\ &= p^{n-r} + (p-1)(p^{n-r}-1) + p-1 \\ &= p^{n-r+1}  \end{aligned}

as required. When r \ge r'+1 we have c > r' in all summands so

\begin{aligned} L &= -(p-1)p^{n-r} + (p-1)\sum_{c=r}^{n-1} (p-1)p^{n-c-1} + p^{r-1}(p-1) \\ &= (p-1) \bigl( -p^{n-r} + (p^{n-r}-1) + 1 \bigr)  \\ &= 0. \end{aligned}

When r \le r'-1 the first non-zero summand occurs for c=r'-1 so we have

\begin{aligned} L  &= - (p-1)p^{n-r'} + (p-1) \sum_{c=r'}^{n-1} (p-1)p^{n-c-1} + p^{r-1}(p-1) \\ &= (p-1) \bigl( -p^{n-r'} + (p^{n-r'}-1) + 1 \bigr) \\ &=0. \end{aligned}

Now taking determinants, using that \det R(p^n)^\circ = \det R(p^n) (the matrices are conjugate by the matrix with 1s on its antidiagonal and 0s elsewhere), we get 1 / \det R(p^n) = (p^n)^{n+1} \det R(p^n). Hence \det R(p^n) = p^{n(n+1)/2}, as required.

Jordan block matrices.

Let J be the m \times m unipotent upper-triangular Jordan block matrix over \mathbb{F}_p. We have (J-I)^m = 0, hence (J-I)^{p^r} = 0 whenever p^r \ge m. On the other hand if k \le m-1 then J^k has a 1 in position (1,k). Therefore J^{p^r} = I if and only if p^r \ge m. We also need a result on the relative trace matrix I + J + \cdots + J^{p^{r}-1}. Note that \binom{p}{k} is divisible by p for k \in \{1,\ldots, p-1\}. (For instance, a p-cycle acts freely on the set of k-subsets of \{1,\ldots, p\}.) An easy inductive argument using \binom{p-1}{k-1} + \binom{p-1}{k} = \binom{p}{k} shows that \binom{p-1}{k} \equiv (-1)^k mod p for all k. Now Lucas’ Theorem implies that \binom{p^r-1}{k} \equiv (-1)^k mod p. Hence

(J-I)^{p^r-1} = I + J + \cdots + J^{p^r-1}.

It follows that the right-hand side is 0 if and only if p^r > m.

Regular abelian subgroups of permutation groups

March 15, 2017

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.