TITLE: Finite groups in which every character has real values: grading the representations
QUESTION [14 upvotes]: Let $G$ be a finite group. Then the irreducible complex representations of $G$ come in three sorts: real, complex and symplectic=quaternionic. The type of an irreducible character $\chi$ can be read of from the Frobenius-Schur indicator 
$$ s_2(\chi) = \frac{1}{|G|}\sum_{g\in G} \chi( g^2 ) \in \{ 1,0,-1 \}. $$
Now the following seems to be true (and has been used by Noah Snyder in his interesting answer to  another question) but I can't see why:
Suppose all characters of $G$ have real values. (Equivalently, every element of $G$ is conjugate to its inverse.) Then it seems that the Frobenius-Schur indicator defines a grading of the irreducible representations, and thus of the character ring. This means that if $\chi$ and $\psi$ are irreducible characters with $s_2 (\chi) = s_2(\psi)$, then all the irreducible constitutents of $\chi\psi$ have indicator $1$, and if $s_2(\chi) = -s_2(\psi)$, then all constituents of $\chi\psi$ have indicator $-1$. Why is this actually true?  Of course, for example in the first case, $\chi\psi$ is afforded by a real representation, so the symplectic representations must occur with even multiplicity. But why can they not occur at all?  
Moreover, I would like to know if this generalizes to other fields than $\mathbb{R}$, using elements of a Brauer group instead of the Frobenius-Schur indicator.  
EDIT: The statement turned out to be wrong  in general (see below), so the original question is in some sense obsolete. A more appropriate questions would have been why this Frobenius-Schur indicator grading is there in some (many?) cases.

REPLY [3 votes]: Here is a variation of the question which is true and generalizes readily to the Brauer group.
If $G$ and $H$ are two groups with two complex irreducible representations $V$ and $W$, then $V \otimes_{\mathbb{C}} W$ is an irrep of $G \times H$, and the Schur-Frobenius indicators multiply.
For the proof, I'll switch to the generalization that works for any Brauer group.  Let $F$ be a field of characteristic zero and let $V$ and $W$ be two irreps of $G$ and $H$ over the field $F$.  Then $V$ and $W$ might each represent an element of the Brauer group of $F$, because by Schur's lemma each of $\mathrm{End}_G(V)$ and $\mathrm{End}_H(W)$ is a division ring.  I say "might" because it is not necessarily true that the centers of these division rings are $F$.  If so, then $V \otimes_F W$ does not quite have to be $(G \times H)$-irreducible, but its summands land in the expected place in the Brauer group, using the equation
$$\mathrm{End}_{G \times H}(V \otimes_F W) \cong \mathrm{End}_G(V) \otimes_F \mathrm{End}_H(V).$$
Taking the case $F = \mathbb{R}$, you get a relation that is equivalent to the previous paragraph.

Here also is a remark about Noah Snyder's answer to the other question.  He points out that the representation ring of $G$ is graded by the character group $Z(G)^*$ of the center of $G$.  It sometimes so happens, when the characters of $G$ are all real, that the Schur-Frobenius indicator is given by a group homomorphism from $Z(G)^*$ to $\{\pm 1\}$.  If so, then of course the indicator gives you a reduced grading of the representation ring of $G$.  So, it would be interesting to know which finite or compact groups have this property, or an analogous property for another Brauer group.  I think (not quite sure) that all compact simple Lie groups with real characters are examples.