Burnside’s method

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.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: