let G be a finite group. Suppose C is the set of conjugacy classes of G and R is the set of (equivalence classes of) irreducible representations of G over the complex numbers.

The automorphism group of G has a natural action on C and also on R (we can make both of these left actions). My questions:

  1. Under what conditions are C and R equivalent as $\operatorname{Aut}(G)$-sets? This is definitely true, for instance, if every automorphism is inner, if the outer automorphism group of G is cyclic (it then follows from Brauer's permutation lemma) and it is also true if the quotient of the automorphism group by the group of class-preserving automorphisms is cyclic (again by Brauer's permutation lemma). But it also seems to be true in a number of other cases, such as the quaternion group, where the outer automorphism group is a symmetric group of degree three.
  2. A weaker condition: under what conditions are the orbit sizes under $\operatorname{Aut}(G)$ for C and R the same?
  • $\begingroup$ Can you give an example of a group for which this condition ($\operatorname{Aut}(G)$-equivalence of $C$ and $R$) does not hold? $\endgroup$ Commented Apr 16, 2010 at 19:32
  • $\begingroup$ Not offhand, but I'm somehow convinced they're not equivalent, or this would be another grand theorem of representation theory. $\endgroup$
    – Vipul Naik
    Commented Apr 16, 2010 at 19:46
  • $\begingroup$ The reason for my skepticism is that while we know how to set up an explicit "natural" bijection between conjugacy classes and irreducible representations in only a very few special cases (such as the symmetric and alternating groups, all of which satisfy your sufficient conditions), I've never heard it said anywhere that it's known that such an explicit bijection does not exist in general. Of course, this line of reasoning is assuming that such bijections would necessarily be $\operatorname{Aut}(G)$-invariant; I'm not certain this is even true for the cases of $S_n$ and $A_n$. $\endgroup$ Commented Apr 16, 2010 at 19:58
  • $\begingroup$ Prof Alperin told me that no such bijection exists, though I don't remember how he made the statement precise. $\endgroup$
    – Vipul Naik
    Commented Apr 16, 2010 at 20:16
  • $\begingroup$ For $S_n$, there is a bijection because the irreps and conjugacy classes are both indexed by partitions of <em>n</em>. $\endgroup$
    – Vipul Naik
    Commented Apr 16, 2010 at 20:17

2 Answers 2


I think that an example of non-equivalent permutation sets is given by $G=(\mathbb Z/p\mathbb Z)^n$ for $n>2$ (and $p$ a prime). Then the automorphism group is $\mathrm{GL}_n(\mathbb Z/p\mathbb Z)$, the conjugacy classes are in natural bijection with $G$ and the set of irreducible representations are in bijection with the dual group (or dual $\mathbb Z/p\mathbb Z$-vector space). In both cases there are only two orbits, one of length $1$ (the identity element and the trivial representation respectively). The stabilisers for elements in the non-trivial orbits are not conjugate: Mapping to $\mathrm{PGL}_n(\mathbb Z/p\mathbb Z)$ map these two kinds of stabilisers two non-conjugate parabolic subgroups (stabilisers of lines resp. of hyperplanes).

  • 1
    $\begingroup$ Thanks! So it's the non-naturality of the isomorphism between an abelian group and its dual that I think is coming into play here. This is a great example of a case where (2) holds but (1) doesn't. $\endgroup$
    – Vipul Naik
    Commented Apr 17, 2010 at 14:09
  • 3
    $\begingroup$ This example explains why we should expect something in-between (1) and (2) to hold: Is there always an automorphism $a:Aut(G)\to Aut(G)$ and a bijection $b:C\to R$ such that $b(g\cdot c)=a(g)\cdot b(c)$? $\endgroup$ Commented Oct 28, 2010 at 4:23

By Brauer's permutation lemma, the permutation characters are always equal, but the representations need not be isomorphic. For instance, the non-abelian group of order 27 and exponent 9 provides an example. One condition for an equivalence for subgroups of the automorphism group is given in Isaacs's Character Theory textbook as theorem 13.24 on page 230–231:

If S is a solvable subgroup of Aut(G), and gcd(|S|,|G|)=1, then the permutation representations of S on Irr(G) and Cl(G) are isomorphic.

This will rarely directly answer your question as Aut(G) and G usually have common prime divisors, but perhaps the ideas will be useful to you. In particular, it describes a strengthening of your #2 which implies #1.

Let me know if you would like GAP code to verify the order 27 example. The action on classes has orbits of sizes 1, 1, 1, 2, 6 and the action on the irreducibles has orbits of sizes 1, 2, 2, 3, 3.

GAP code to check permutation isomorphism:

OnCharactersByGroupAutomorphism := function( pnt, act )
  return Character( UnderlyingCharacterTable( pnt ),
  pnt{FusionConjugacyClasses(act^-1)} );
OnCBGA := OnCharactersByGroupAutomorphism;;

g := ExtraspecialGroup(27,9);;
a := AutomorphismGroup(g);;
gensIrr := List( GeneratorsOfGroup(a), f ->
  PermListList( Irr(g), List( Irr(g), chi -> OnCBGA( chi, f ) ) ) );
gensCcl := List( GeneratorsOfGroup(a), f ->
  PermList( FusionConjugacyClasses(f) ) );
# perm iso?
fail <> RepresentativeAction( SymmetricGroup( NrConjugacyClasses( g ) ),
  gensCcl, gensIrr, OnTuples );

Some of what you asked for might be more along the lines of asking if the permutation groups generated by gensIrr and gensCcl are conjugate, so I chose an example where even the images are not conjugate. The example given below of G=2×2×2 is the smallest if you only want strict permutation (non-)isomorphism.

  • $\begingroup$ I'd be grateful if you could provide the GAP code, or some fragment of it. Thanks for the helpful answer. $\endgroup$
    – Vipul Naik
    Commented Apr 17, 2010 at 14:02
  • $\begingroup$ Very helpful answer ! May I ask you: the theorem you cite talks about "Aut" but actually only "Out" acts, is there some modification for "Out" ? $\endgroup$ Commented Sep 11, 2012 at 6:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.