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Dan Petersen
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You can use the pre-$\lambda$ ring identity $$ \big(\sum_n [\mathrm{Sym}^n(X)]t^n\big)\big( \sum_k [\wedge^k(X)](-t)^k\big) = 1.$$$$ \big(\sum_n [\mathrm{Sym}^n(X)]t^n\big)\big( \sum_k \lambda^k[X](-t)^k\big) = 1.$$ So it is enough to observeshow that the generating series for the exterior powers is a polynomial, for any finite $G$-set $X$. Here $\wedge^k(X)$There is thean explicit formula for $G$-set$\lambda^k[X]$ in a paper of $k$-element subsets ofRökaeus, from which it in particular follows that $X$$\lambda^k[X]=0$ if $k>|X|$.

You can use the pre-$\lambda$ ring identity $$ \big(\sum_n [\mathrm{Sym}^n(X)]t^n\big)\big( \sum_k [\wedge^k(X)](-t)^k\big) = 1.$$ So it is enough to observe that the generating series for the exterior powers is a polynomial, for any finite $G$-set $X$. Here $\wedge^k(X)$ is the $G$-set of $k$-element subsets of $X$.

You can use the pre-$\lambda$ ring identity $$ \big(\sum_n [\mathrm{Sym}^n(X)]t^n\big)\big( \sum_k \lambda^k[X](-t)^k\big) = 1.$$ So it is enough to show that the generating series for the exterior powers is a polynomial, for any finite $G$-set $X$. There is an explicit formula for $\lambda^k[X]$ in a paper of Rökaeus, from which it in particular follows that $\lambda^k[X]=0$ if $k>|X|$.

Source Link
Dan Petersen
  • 40.2k
  • 2
  • 114
  • 201

You can use the pre-$\lambda$ ring identity $$ \big(\sum_n [\mathrm{Sym}^n(X)]t^n\big)\big( \sum_k [\wedge^k(X)](-t)^k\big) = 1.$$ So it is enough to observe that the generating series for the exterior powers is a polynomial, for any finite $G$-set $X$. Here $\wedge^k(X)$ is the $G$-set of $k$-element subsets of $X$.