The proof uses the decomposition of ${{\mathbb R}}G$ as an ${{\mathbb R}}$-algebra in terms of the decomposition of ${{\mathbb C}}G$ as a ${{\mathbb C}}$-algebra and so we need to recall how this comes about. Remember that $X({{\mathbb C}}G)$ stands for the set of characters of irreducible ${{\mathbb C}}G$-modules, that for every $\chi \in X({{\mathbb C}}G)$ we have chosen a representative ${{\mathbb C}}G$-module $V_\chi$, and that the obvious map $$\textstyle {{\mathbb C}}G\to \prod_{\chi\in X({{\mathbb C}}G)} {\operatorname{End}}(V_\chi)$$ is an isomorphism of ${{\mathbb C}}$-algebras. ‘Taking the dual’ defines an involution $\chi\mapsto \chi^*$ in $X({{\mathbb C}}G)$. Then for all $\chi, \mu \in X({{\mathbb C}}G)$, the space of $G$-invariant bilinear forms $V_\chi\times V_{\mu}\to {{\mathbb C}}$ is nonzero unless $\mu=\chi^*$, in which case it is generated by a nondegenerate pairing $\alpha_\chi: V_\chi\times V_{\chi^*}\to {{\mathbb C}}$. The fixed point set of this involution decomposes into the orthogonal characters $X({{\mathbb C}}G)_+$ (the $\chi$ for which $\alpha_\chi$ is symmetric) and the symplectic characters $X({{\mathbb C}}G)_-$ (the $\chi$ for which $\alpha_\chi$ is anti-symmetric). Let $X({{\mathbb C}}G)_o\subset X({{\mathbb C}}G)$ be a system of representatives for the free orbits, so that $X({{\mathbb C}}G)$ is the disjoint union of $X({{\mathbb C}}G)_+$, $X({{\mathbb C}}G)_-$, $X({{\mathbb C}}G)_o$ and $X({{\mathbb C}}G)_o^*$. Choose a $G$-invariant inner product 