$G$-invariant symmetric, nondegenerate form is unique up to scalar

Question

Let $V$ be a f.d. representation of a finite group $G$ over a field $F$. A standard argument shows there is a $G$-invariant, symmetric, nondegenerate bilinear form on $V$. If $(-,-)$ is any such form, not necessarily invariant, define $$\langle v,w\rangle := \sum\limits_{x \in G} (xv,xw)$$ I remember proving for homework that any two such $G$-invariant forms differ by a scalar. I think I did something with the adjoint representation on $V^{\ast}$ I tried to reproduce what I did before but I can't seem to remember it. Would someone kindly give me a reference or a proof?

Answer 1

If $\rho:G\to \mathrm{GL}(V)$ is irreducible and $F$ is algebraically closed (and $\lvert G\rvert$ is not divisible by the characteristic of $F$), then $\dim\hom_G(V,V^*)$ is either zero or one since $V^*$ is also irreducible. Using the fact that a $G$-invariant bilinear form is an element of $\hom_G(V,V^*)$, from $v\mapsto(w\mapsto (v,w))$, then since it is nondegenerate, $\hom_G(V,V^*)$ is one-dimensional, and all such forms are scalar multiples of one another.

Or, if $F$ is a field where symmetric matrices are diagonalizable, let $A$ and $B$ be the matrices of two nondegenerate $G$-invariant bilinear forms. Then $x^TA(By)=(By)^TAx=y^TB^TAx=y^TBAx$, and since $AB$ is symmetric, $AB=BA$, hence $A$ and $B$ are simultaneously diagonalizable. This implies there is some constant $c$ (a ratio of eigenvalues) such that $A-cB$ is a degenerate symmetric $G$-invariant form. Because $G$-invariant forms are elements of $\hom_G(V,V^*)$, by Schur's lemma $A-cB=0$, so $A=cB$.