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Let $\rho:G\rightarrow GL(V)$ be a complex representation. For each $n$, let $\chi_{\text{Sym}^n}$ be the character of the n-th symmetric power of $V$. Prove for each $g\in G$, $$\sum_{i=0}^\infty \chi_{\text{Sym}^n}(g)t^n=\frac{1}{\det(I-t\rho(g))}$$. I find the Molien's theorem, but there is a summation on the right hand side.

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Let $\lambda_1, \cdots, \lambda_m$ be the eigenvalues of $\rho(g)$, and $h_i$ the $i$-th complete homogeneous symmetric polynomial of $\lambda_1, \cdots, \lambda_m$. Then $\chi_{\text{Sym}^n(g)}=h_n$, and

\begin{align*} \frac{1}{\det(I-t\rho(g))}&=\frac{1}{\prod_{i=1}^m(1-t\lambda_i)}\\ &=\prod_{i=1}^m(1+t\lambda_i+t^2\lambda_i^2+\cdots)\\ &=1+h_1t+h_2t^2+\cdots\\ &=1+\chi_{\text{Sym}(g)}t+\chi_{\text{Sym}^2(g)}t^2+\cdots \end{align*}

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