The complex irreps of a finite group come in three types: self-dual by a symmetric form, self-dual by a symplectic form, and not self-dual at all. In the first two cases, the character is real-valued, and in the third it is sometimes only complex-valued. The cases can be distinguished by the value of the Schur indicator $\frac{1}{|G|} \sum_g \chi(g^2)$, necessarily $1$, $-1$, or $0$. They correspond to the cases that the representation is the complexification of a real one, the forgetful version of a quaternionic representation, or neither.

A conjugacy class $[g]$ is called "real" if all characters take real values on it, or equivalently, if $g\sim g^{-1}$. I vaguely recall the number of real conjugacy classes being equal to the number of real irreps.

  1. Do I remember that correctly?

  2. Can one split the real conjugacy classes into two types, "symmetric" vs. "symplectic"?

With #1 now granted, a criterion for a "good answer" would be that the number of symmetric real conjugacy classes should equal the number of symmetrically self-dual irreps.

(I don't have any application in mind; it's just bothered me off and on for a long time.)

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    $\begingroup$ Two types? or three? $\endgroup$ Nov 22, 2010 at 3:57
  • $\begingroup$ Perhaps whether the permutation of the conjugacy class by inversion is even or odd? $\endgroup$
    – Ben Webster
    Nov 22, 2010 at 4:30
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    $\begingroup$ Theorem 23.1 in James and Liebeck "Representation of Finite Groups" states that the number of real conjugacy classes is equal to the number of real irreps. I don't know about the second question though $\endgroup$ Nov 22, 2010 at 4:31
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    $\begingroup$ +1, neat question! @Ben, I don't think the sign of the permutation by inversion is correct. Under your definition, the conjugacy class of $(1 2 3)$ in $S_3$ would be symplectic. But, as everyone knows, all the conjugacy classes of $S_n$ should be symmetric, just like all the characters! $\endgroup$ Nov 22, 2010 at 7:58
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    $\begingroup$ @Allen: Maybe you will give us more of a hint of what you expect: the group $Q_8$ of quaternions has $4$ symmetrically self dual (real) representations and $1$ skew-symmetrically self-dual (quaternionic) representation, so we should expect exactly $1$ "quaternionic" conj. class. Given that the labeling should be stable by automorphisms, this special class must be either $1$ or $-1$. Which do you suppose it is? $\endgroup$ Nov 22, 2010 at 14:47

3 Answers 3


It's a great question! Disappointingly, I think the answer to (2) is No :

The only restriction on a `good' division into "symmetric" vs. "symplectic" conjugacy classes that I can see is that it should be intrinsic, depending only on $G$ and the class up to isomorphism. (You don't just want to split the self-dual classes randomly, right?) This means that the division must be preserved by all outer automorphisms of $G$, and this is what I'll use to construct a counterexample. Let me know if I got this wrong.

The group

My $G$ is $C_{11}\rtimes (C_4\times C_2\times C_2)$, with $C_2\times C_2\times C_2$ acting trivially on $C_{11}=\langle x\rangle$, and the generator of $C_4$ acting by $x\mapsto x^{-1}$. In Magma, this is G:=SmallGroup(176,35), and it has a huge group of outer automorphisms $C_5\times((C_2\times C_2\times C_2)\rtimes S_4)$, Magma's OuterFPGroup(AutomorphismGroup(G)). The reason for $C_5$ is that $x$ is only conjugate to $x,x^{-1}$ in $C_{11}\triangleleft G$, but there there are 5 pairs of possible generators like that in $C_{11}$, indistinguishable from each other; the other factor of $Out\ G$ is $Aut(C_2\times C_2\times C_4)$, all of these guys commute with the action.

The representations

The group has 28 orthogonal, 20 symplectic and 8 non-self-dual representations, according to Magma.

The conjugacy classes

There are 1+7+8+5+35=56 conjugacy classes, of elements of order 1,2,4,11,22 respectively. The elements of order 4 are (clearly) not conjugate to their inverses, so these 8 classes account for the 8 non-self-dual representations. We are interested in splitting the other 48 classes into two groups, 28 'orthogonal' and 20 'symplectic'.

The catch

The problem is that the way $Out\ G$ acts on the 35 classes of elements of order 22, it has two orbits according to Magma - one with 30 classes and one with 5. (I think I can see that these numbers must be multiples of 5 without Magma's help, but I don't see the full splitting at the moment; I can insert the Magma code if you guys want it.) Anyway, if I am correct, these 30 classes are indistinguishable from one another, so they must all be either 'orthogonal' or 'symplectic'. So a canonical splitting into 28 and 20 cannot exist.

Edit: However, as Jack Schmidt points out (see comment below), it is possible to predict the number of symplectic representations for this group!

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    $\begingroup$ G has 40 weakly real classes, 8 strongly real classes. Since it is 2-nilpotent with Abelian Sylow 2-subgroup, it has exactly 8+40/2 = 28 characters of indicator +1. In other words, the weakly real classes only count in pairs. $\endgroup$ Nov 22, 2010 at 19:16

A standard strengthening of "real element" is "strongly real element". An element is strongly real if it is conjugate to its inverse by an involution, or equivalently, if it is a product of involutions (equivalently, it sits nicely inside a dihedral group).

The quaternion group of order 8 shows that this strengthening is distinct: the only strongly real element is the identity, but Q8 has 4 representations with Frobenius–Schur indicator +1.

However, in:

Gow, R. "Real-valued and 2-rational group characters." J. Algebra 61 (1979), no. 2, 388–413 MR2222410 DOI:10.1016/0021-8693(79)90288-6

some inequalities relating the two ideas are given as well as a few reasonably strong results. If the Sylow 2-subgroups are dihedral or large enough semi-dihedral, then a quaternionic representation exists iff a non-strongly real but real element of odd order exists (theorems are spread out in the paper). In the case of a 2-nilpotent group (the last section) more precise relationships are given, with the number of quaternionic reps being a mixture of the number of strongly and weakly real elements.

Beware of wanting to generalize the real case too broadly. An F-rational character is a C-irreducible character whose values are in F. An F-rational element is an element conjugate to the correct powers given the (cyclotomic) Galois group of C/F. In p-groups for odd p, the F-rational classes are 1–1 with F-rational characters, but not in general for 2-groups or groups of odd order.

  • $\begingroup$ Why isn't -1 also a strongly real element? $\endgroup$ Mar 11, 2011 at 23:07
  • $\begingroup$ @DavidLHarden: −1 in Q8 is also strongly real $\endgroup$ Mar 12, 2011 at 16:41
  • $\begingroup$ Flip side of my original question: is there a corresponding notion of "strongly real character"? $\endgroup$ Feb 21, 2023 at 19:44

The recent paper, "A solution to Brauer's Problem 14" by John Murray and Benjamin Sambale, – J Algebra 621 (2023) 87-91, is relevant to this question. It gives a way of counting the number of characters with Frobenius-Schur indicator +1 just by counting solutions of $g_1^2\dotsm g_n^2=1$ with $g_1,\dotsc,g_n \in G$.


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