Brauer's permutation lemma -- extending to some other finite groups? - MathOverflow

Brauer's permutation lemma -- extending to some other finite groups?

2010-04-13T13:11:12Z

Brauer's permutation lemma states that any two permutation matrices are conjugate in $GL(n,\mathbb{C})$ if and only if they are conjugate in the symmetric group, i.e., they have the same cycle type (we can replace $\mathbb{C}$ by any field of characteristic zero).

Another way of stating it is that any two permutation representations of a finite cyclic group that are conjugate as linear representations in characteristic zero are in fact equivalent as permutation representations.

My question: for what classes of finite groups does this result hold? i.e., for what classes of finite groups is it true that any two permutation representations that are conjugateas linear representations in characteristic zero are in fact equivalent as permutation representations?

I think some counterexamples involving Mathieu groups exist, so this is not true for all finite groups. What I was looking for are theorems of the form that it is true for all groups satisfying some well-studied property or in some well-identified class.

One application of this would be as follows: suppose G is a finite group such that the quotient of the automorphism group of G by the group of class-preserving automorphisms of G satisfies the above condition. (Class-preserving automorphisms are automorphisms that preserve each conjugacy class; for a finite group, this is equivalent to preserving all the characters). Then, the orbit sizes under the action of the automorphism group of G are the same for the set of conjugacy classes and the set of (equivalence classes of) irreducible representations.

Answer by Ben Webster for Brauer's permutation lemma -- extending to some other finite groups?

2010-04-13T13:49:34Z

There are many group actions on sets which are linearly equivalent but not equivalent as actions. In fact, every group other than the cyclic group has one. This follows from some easy linear algebra:

the number of irreducible reps over $\mathbb{Q}$ is the number of conjugacy classes of cyclic subgroups of $G$, (EDIT:there are at most this many since any two elements which generate conjugate cyclic groups have the same character in a rational representation; on the other hand, the characters of the inductions of the trivial from any set of cyclic groups, no two of which are conjugate, are linearly independent, so there are at least this many) and
the number of non-isomorphic transitive G-sets is the number of conjugacy classes of subgroups.

Thus, there must be an integer valued linear combination of transitive actions which has trivial character. Moving all the actions with negative coefficients to the other side of the equality, we get two different actions with the same character, and thus isomorphic representations.

I actually wrote a paper about this a few years back, which I think is a reasonable starting place for the subject, which actually has quite a long history, and a reasonably extensive literature.

Answer by anonymous for Brauer's permutation lemma -- extending to some other finite groups?

2010-04-13T14:10:21Z

Let $G$ be a finite subgroup of $GL(V)$. I claim that the following are equivalent:

$(1)$ Any two elements of $G$ which are $GL(V)$ conjugate are also $G$ conjugate.

$(2)$ The representation ring $\mathbb{Q} \otimes \mathrm{Rep}(G)$ is spanned by representations $S^{\lambda}(V)$, where $S^{\lambda}$ ranges over all Schur functors.

Proofs: For $g$ in $G$, let the eigenvalues of $g$ be $e^{2 \pi i q_k(g)}$ , where $(q_1(g), q_2(g), \ldots, q_n(g))$ is a multiset of elements of $\mathbb{Q}/\mathbb{Z}$. Write $Q(g)$ for this multiset. So $g$ and $g'$ are $GL(V)$ conjugate if and only if $Q(g) = Q(g')$. When necessary, we will write $Q_W(g)$ for the analogous construction for some other representation $W$.

If $Q(g)=Q(g')$ then $Q(g^k) = k Q(g) = k Q(g') = Q((g')^k)$. So, for every Adams operator $\psi^k$, the actions of $g$ and $g'$ on the virtual representation $\psi^k(V)$ have the same eigenvalues and, in particular, the same trace. Also, $Q_{V_1 \oplus V_2}(g)$ and $Q_{V_1 \otimes V_2}(g)$ are determined by $Q_{V_1}(g)$ and $Q_{V_2}(g)$ . So, if $Q(g)=Q(g')$ then the linera functions $Tr(g)$ and $Tr(g')$ are equal on the sub-$\Lambda$-ring of $\mathbb{Q} \otimes \mathrm{Rep}(G)$ generated by $V$.

Assume condition 2, and let $Q(g) = Q(g')$. Then $Tr(g)$ and $Tr(g')$ are equal on all of $\mathbb{Q} \otimes \mathrm{Rep}(G)$, so $g$ and $g'$ are $G$ conjugate.

Conversely, assume condition 1. Fix any $h \in G$. Let $S$ be the sub-$\Lambda$-ring of $\mathbb{Q} \otimes \mathrm{Rep}(G)$ generated by $V$. We will construct a virtual representation $W$ in $\mathbb{C} \otimes S$ such that $Tr_W(g)$ is $0$ for $g$ not conjugate to $h$ and $1$ for $g$ conjugate to $h$. Taking such linear combinations of such linear functionals, we can clearly generate all class functions on $G$, so $S$ must be the whole of $\mathbb{Q} \otimes \mathrm{Rep}(G)$.

By condition (1), $Q(h)$ is different from $Q(g)$ for any $g$ not conjugate to $h$. Therefore, we can find a symmetric polynomial $F$, with coefficients in $\mathbb{C}$, which vanishes at $e^{2 \pi i Q(g)}$ for $g$ not conjugate to $h$, but does not vanish at $e^{2 \pi i Q(h)}$. (Here $e^{2 \pi i Q(g)}$ is a point of $\mathbb{C}^n$, which should be considered as defined only up to the action of $S_n$.) Using the standard relation between symmetric polynomials and $\Lambda$-ring operations, $F(V)$ is the desired $W$.

Answer by Bugs Bunny for Brauer's permutation lemma -- extending to some other finite groups?

2010-04-13T21:38:40Z

You may be able to salvage something by looking at the Bursnide ring of your group. The isomorphism class of the permutation representation of a set $X$ is governed by products $[X]e$ with some of the idempotents and you need to include missing idempotents (of the Burnside ring) to have a version of Brauer's permutation lemma for any group...