 Solution

Let $T(v\otimes w)=w\otimes v$ on the tensor product $V\otimes V$. The linear map $R(g)\otimes R(g)$ commutes with $T$, so preserves its $+1$ and $-1$ eigenspaces, namely the symmetric square $S^{2}V$ and the exterior square ${\wedge }^{2}V$. Restricting the tensor product representation to these invariant subspaces defines $S^{2}R$ and ${\wedge }^{2}R$. The representation identity follows by restricting $(R(g)\otimes R(g))(R(h)\otimes R(h))=R(gh)\otimes R(gh)$.

For a finite group, $g$ has finite order $m$, so $R(g{)}^m=I$. Over the complex numbers, the polynomial $t^m-1$ has distinct roots; hence $R(g)$ is diagonalizable. Choose an eigenbasis $v_1,\dots ,v_n$ with eigenvalues ${\lambda }_1,\dots ,{\lambda }_n$. A basis of the symmetric square consists of $v_i\otimes v_i$ and $v_i\otimes v_j+v_j\otimes v_i$ for $i<j$. A basis of the exterior square consists of $v_i\otimes v_j-v_j\otimes v_i$ for $i<j$. Their eigenvalues are respectively ${\lambda }_i^2,{\lambda }_i{\lambda }_j$ and ${\lambda }_i{\lambda }_j$. Taking traces,But $\chi (g{)}^2={\sum }_i{\lambda }_i^2+2{\sum }_{i<j}{\lambda }_i{\lambda }_j$ and $\chi (g^2)={\sum }_i{\lambda }_i^2$. Thus the characters areAt $g=e$, these also give the dimensions $n(n+1)/2$ and $n(n-1)/2$.