Burnside proved in 1901 that if is an odd prime then a permutation group containing a regular subgroup isomorphic to 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 is a B-group for any prime and any , 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 -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, acts primitively on , and has a regular subgroup isomorphic to . 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 be a primitive th root of unity. Define
for . Define a subset of to be null if there exists and distinct for and such that mod for each and and
Proposition 6.2 Let where is not divisible by . Let . Then
if and only if either
- is null; or
- where is a null set, the are distinct elements of and mod for each .
Proof. Since the minimum polynomial of is
we have . Since , we have . It follows that if is a null set. (For the second equality, note the contributions from the for fixed combine to give .) For (2) we have , and . This proves the ‘if’ direction.
Conversely, by Lemma 2.1 in the paper, is a union of some of some of the sets . There exists a unique subset of and unique for and unique for and such that
We have and . Therefore
has as a root. Since this polynomial has degree at most and the minimal polynomial of is , it follows that the coefficients are constant. Hence
for . If then and is null. Otherwise, taking the previous displayed equation mod we see that , and, moreover, the are constant for . (This holds even if .) Set
We have . Hence, and choosing in any way for such that , we see that is the union of , the sets for and a null set. .
For prime and we define to be the matrix
More generally, if has prime factorization , we define . The rows and columns of are labelled by the divisors of , as indicated below for the case , with and odd prime.
Say that a partition of is coprime if the highest common factor of the numbers in each part is . The aim of the game is to find a non-empty set of rows, say, of such that the subsets of columns (excluding column ) on which the sum of the rows in are equal form a coprime partition of the divisors.
There is an application to B-groups only when is even, in which case we may assume that . Proposition 6.7 in the paper states that in this case, the only way to win this game when , or , where is an odd prime, is to put every single row in . This implies that is a B-group when or . However the result on the game may well hold more generally.
Constant sums for
The only coprime partition of has a singleton part, so the row sums are all equal. By adding to each entry in row for , we obtain the matrix below.
which still has constant sums over the rows in . Let if and let otherwise. Writing numbers in base , the row sums over are, for each column,
(Note in each case there are digits, the most significant corresponding to .) The numbers in the right column are equal, hence all the are equal and , as required. Taking this gives an alternative proof of Proposition 6.7(i).
Proofs of further claims on the Ramanujan matrices
Let denote rotated by a half turn. The following result was stated without proof in the paper; it implies that each is invertible, with inverse . (Update: I later found a much better proof, now outlined in the paper, using a lower-upper decomposition of .)
Proposition. We have and .
Proof. Since , we have for all and
for and . We use this to show that
for . When the first term in each product is , so the left-hand side is the column sum of the -th of ; this is if and otherwise, as required. Now suppose that . Since vanishes when , the left-hand side is
Take out a factor to define . Substitute for , and split off the summand from to get
We must now consider three cases. When we get
as required. When we have in all summands so
When the first non-zero summand occurs for so we have
Now taking determinants, using that (the matrices are conjugate by the matrix with s on its antidiagonal and s elsewhere), we get . Hence , as required.
Jordan block matrices.
Let be the unipotent upper-triangular Jordan block matrix over . We have , hence whenever . On the other hand if then has a in position . Therefore if and only if . We also need a result on the relative trace matrix . Note that is divisible by for . (For instance, a -cycle acts freely on the set of -subsets of .) An easy inductive argument using shows that mod for all . Now Lucas’ Theorem implies that mod . Hence
It follows that the right-hand side is if and only if .