The Permutation Lemma of Richard Brauer

Dear Charlie, I promised to write to you about the "characteristic-free" extension of Richard Brauer's Lemma: if a row permutation and a column permutation have the same effect on a nonsingular matrix, the two permutations must have the same number of cycles of any given length. Differently put: if two permutation matrices are conjugate in the general linear group, they are already conjugate in the symmetric group; or, if two permutation representations of a cyclic group are equivalent as matrix representations, they are also equivalent as permutation representations. The latter form was certainly known to Burnside: as Peter Neumann has kindly reminded me, §217 of (the second edition of) Burnside's book even discusses the failure of this for all noncyclic groups. Both in Burnside and in Brauer, the matrices have complex entries. Although Brauer's footnote on page 934 of Annals 42 (1941) says that "a modification is necessary, if the field is modular, but the Lemma remains valid", I have not been able to adapt his proof (or Burnside's) to the case of nonzero characteristic. The best characteristic-free proof I know runs as follows. For a permutation P (of some finite set X), let (D(P) denote the number of cycles (or orbits) of P, not ignoring the fixed points. If P is regarded as a matrix acting on a vectorspace (with basis X, over any field of any characteristic), the dimension of the subspace formed by the vectors fixed by P is easily seen to be co(P). Thus if Q is another permutation of X, conjugate to P in the automorphism group of this vectorspace, we must have a>(P) = a){Q), and indeed u)(P) = (o(Q) for all k. Let to,(P) denote the number of cycles of P of length /. The Lemma is now a consequence of the following combinatorial fact (which no longer refers to matrices): if P and Q are permutations such that oj(P) = co(Q) for all k, then to,(P) = co^Q) for all /. This must be in print somewhere, but I have not been able to find a reference. To see it, consider first a single cycle C of length / (on a set of / elements): then a>(C) is the greatest common divisor (k, I) of k and /. It follows that a)(P) = £ (k, /)o>/(F) for all k ,