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EDIT: Ben Webster's answer below still doesn't rule The statement turned out the possibility that a symplectic representation occurs twice in a product of two reps of the same kind, and I don't see how to do this.
For an irreducible $\mathbb{C}G$-module $V$, the type can also seen by looking at $\operatorname{End}_{\mathbb{R}G} be wrong in general (V)$: It's $M_2(\mathbb{R})$ if the type is realsee below), and so the quaternions if $V$ is quaternion. In any case, there original question is $\sigma=\sigma_V\in \operatorname{End}_{\mathbb{R}G}(V)$ with $vz\sigma = v\sigma \bar{z}$ for $z\in \mathbb{C}$ and $v\in V$. In the real case, we may assume $\sigma^2 = 1$, and in the quaternion case we may assume $\sigma^2=-1$. On the tensor product $U\otimes V$ of two modules, we have the element $\tau=\sigma_U\otimes \sigma_V$ with similar properties, and $\tau^2=\pm 1$some sense obsolete. I tried to use A more appropriate questions would have been why this for an argument, but without successFrobenius-Schur indicator grading is there in some (many?) cases.

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EDIT: For some hours I thought I had Ben Webster's answer below still doesn't rule out the possibility that a counterexample to this generalizationsymplectic representation occurs twice in a product of two reps of the same kind, but and I don't see how to do thiswas wrong.It seems to be difficult to find a group where all characters have values in a field
For an irreducible $\mathbb{F}$ \mathbb{C}G$-module$V$, the type can also seen by looking at$\operatorname{End}_{\mathbb{R}G} (V)$: It's$M_2(\mathbb{R})$if the type is real, and such that the endomorphism ring of some simple quaternions if$\mathbb{F}G$-module V$ is a division ring other than quaternion. In any case, there is $\mathbb{F}$ or \sigma=\sigma_V\in \operatorname{End}_{\mathbb{R}G}(V)$with$vz\sigma = v\sigma \bar{z}$for$z\in \mathbb{C}$and$v\in V$. In the quaternions over real case, we may assume$\mathbb{F}$.\sigma^2 = 1$, and in the quaternion case we may assume $\sigma^2=-1$. On the tensor product $U\otimes V$ of two modules, we have the element $\tau=\sigma_U\otimes \sigma_V$ with similar properties, and $\tau^2=\pm 1$. I tried to use this for an argument, but without success.

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Edit: NoFor some hours I thought I had a counterexample to this generalization, at least not directlybut this was wrong. Every character of the group$$C_3 \ltimes (C_9\ltimes C_7) = \langle It seems to be difficult to find a , b, c \mid a^3= b^9= c^7 = 1, c^b = c^2, c^a= c, b^a=b^4 \rangle$$has group where all characters have values in the a field $\mathbb{Q}(\sqrt[3]{1}, \sqrt{-7})$, and there are 12 characters with Schur index 3 over this field (\mathbb{F}$and degree 3). These divide into two classes according to their associated element in the Brauer group. Comparing with such that the universal grading endomorphism ring of some simple$\operatorname{Irr}(G)$, that \mathbb{F}G$-module is the grading by $\operatorname{Irr}(\mathbf{Z}(G))$, we see that the associated class in the Brauer group can not define a grading. (Here the center is $\langle b^3 \rangle division ring other than$, and there is also one character of trivial Schur index which restricts to a given nontrivial character of \mathbb{F}$or the center.)quaternions over$\mathbb{F}\$.

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