## 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 $n$th 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 $1$s on its antidiagonal and $0$s 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.

### 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. $\Box$.