Question Is true that if all real conjugacy classes of a finite group are strongly real, then all its real irreducible representations (irreps) are "strongly real" (symmetric)? And vice versa?

Definitions: An irrep is called real if its Frobenius–Schur indicator is either $+1$ or $-1$, but not $0$. An irrep is called "symmetric" ("orthogonal" ("strongly real")) if it has FS-indicator $+1$, which is equivalent to it being possible to realize it over the real numbers.

A conjugacy class is called "real" if it is preserved by $g\mapsto g^{-1}$. It is called "strongly real" if it is a product of two involutions.

Context It is known that the number of real conjugacy classes is the same as the number of real irreps. A. Knutson asked: Are there “real” vs. “quaternionic” conjugacy classes in finite groups?, i.e., is it possible to split "real" conjugacy classes into two subclasses corresponding to the splitting of irreps. Tim Dokchitser answered that no such splitting may exist in general which will be automorphism covariant ….

The question above is weakening A. Knutson's question. It has been asked explicitly in the paper Kulshrestha and Singh - Real elements and Schur indices of a group (MSN, arXiv), where the authors present some positive evidence for the question above. However, their examples do not cover GAP's SmallGroups List.

SubQuestion How difficult is it to check this in GAP? (I just installed it, but never played.)


2 Answers 2


The answer is No. My source is Rod Gow's rather wonderful little paper "Real-valued characters and the Schur index" (MSN). Let me quote from the introduction:

It may happen that all elements of a group are strongly real while the group possesses real-valued characters of Schur index 2. An example is provided by the central product of quaternion and dihedral groups of order 8.

(The Schur index equals $2$ if and only if the Frobenius–Schur indicator equals $-1$.)


On a related question, see Kaur and Kulshrestha - Strongly real special $2$-groups (arXiv, MSN), which also answers the converse question.


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