Brauer's permutation lemma states that any two permutation matrices are conjugate in $\mathrm{GL}(n,\mathbb{C})$ 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.


3 Answers 3


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.

  • 1
    $\begingroup$ Thanks! I now realize that this is equivalent to the question of whether we can have subgroups of S_n that are conjugate in GL_n but not in S_n. Based on your construction applied to the Klein four-group, I see that S_4 has two Klein four-subgroups that are conjugate in GL_4 but not in S_4. $\endgroup$
    – Vipul Naik
    Apr 13, 2010 at 15:07
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    $\begingroup$ Sorry! For the Klein four-group, we need to take n = 6. The two Klein four-subgroups of S_6 that are not conjugate in S_6 but are conjugate in GL_6 are the subgroups <(1,2)(3,4), (1,3)(2,4)> and <(1,2)(3,4), (1,2)(5,6)>. $\endgroup$
    – Vipul Naik
    Apr 13, 2010 at 15:21
  • $\begingroup$ Right. This actually an interesting example since the Klein 4-group has the largest (compared to the order of the group) minimal pair (exactly as you described) at 3/2 times the order. $\endgroup$
    – Ben Webster
    Apr 13, 2010 at 16:42
  • $\begingroup$ Why "the number of irreducible reps over Q is the number of conjugacy classes of cyclic subgroups of G," ? $\endgroup$ Sep 11, 2012 at 19:26
  • $\begingroup$ I added an explanation. In general, I think for this tangential a question to this old of an answer, asking a new question makes more sense. $\endgroup$
    – Ben Webster
    Sep 12, 2012 at 18:29

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 $W$.

  • $\begingroup$ Thanks! Ben Webster's answer was sufficient for my current purpose, but your approach should be helpful for related applications. $\endgroup$
    – Vipul Naik
    Apr 13, 2010 at 15:17
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    $\begingroup$ I like how we had completely different (but both interesting) interpretations of what "extend" meant. $\endgroup$
    – Ben Webster
    Apr 13, 2010 at 16:44

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...


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