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](https://mathoverflow.net/questions/42646/square-roots-of-elements-in-a-finite-group-and-representation-theory)) 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. 



Finite groups in which every character has real values: grading the representations
As Zoltan suspected in his answer, the statement in my question is not true. I have now found counterexamples: Let $q$ be an odd prime power. Then $SL(2,q)$ contains a group isomorphic to the quaternion group $Q\_8$, which yields a (semiregular) action of $Q\_8$ on $V=(\mathbb{F}\_q)^2$. 
Let $G = V \rtimes Q\_8$ be the semidirect product. All faithful irreps of $G$ have degree $8$
(they are induced from nontrivial linear characters of $V$), and are of real type. 
The other irreps have $V$ in its kernel and come from irreps of $Q\_8$. In particular, all irreps of $G$ are self-dual, and exactly one is of symplectic type, of degree $2$. The tensor product of a faithful irrep with itself contains the symplectic irrep as summand.
Note that $\mathbf{Z}(G)=1$, so there's no nontrivial grading of the irreps of $G$.  

Taking $q=3$ in the above yields a counterexample of order $72$, and I have to admit that I used GAP to find it. For the record, lets note that there are also four groups of order $64$, where all characters are real valued, but the FS-indicator doesn't define a grading of the irreps. 
These are the groups with identifiers [64, 218], [ 64, 224 ], [ 64, 243 ], [ 64, 245 ] in the SmallGroups library of GAP.


