$ G $-invariant symmetric bilinear map and trivial subrepresentation of symmetric square

Question

Let $\pi:G\rightarrow \text{GL}(V)$ be a finite-dimnesional complex representation of finite group $ G $. If there is a non-zero $ G $-invariant symmetric bilinear form on $ V $, then the symmetric square $\text{Sym}^2(V)$ must contain the trivial subrepresentaion. I don't know how to get the trivial subrepresentation. I try to use the universal property of tensor product to get a linear function $\text{Sym}^2(V)\rightarrow\mathbb{C}$ which is invariant under the $ G $-action, but I dont't know what to do next. Any help or hit would be highly appreciated!

Answer 1

Let $b: V \times V \rightarrow \mathbb{C}$ be a $G$-invariant symmetric bilinear form on $V$.

1. Define $f: S^2(V) \rightarrow \mathbb{C}$ by $f(vw) = b(v,w)$ for all $v, w \in V$. (Such a map exists since $b$ is symmetric bilinear.)

2. Check that $f$ is a surjective homomorphism of $\mathbb{C}[G]$-modules. (In other words, $G$-invariant.)

3. Conclude that $S^2(V)$ has a trivial submodule.