# $k[V]^G = \widetilde{A}$ where $\widetilde{A}$ is the normalization of $A$

Let $V$ be a finite dimensional vector space over $k =\overline{k}$ and let $G$ be a subgroup of $GL(V)$ so that $k[V]^G$ is finitely generated. Let $A$ be a subring of $k[V]^G$ that is finitely generated, graded, and separated. Then the normalization $\widetilde{A}$ of $A$ is equal to $k[V]^G$.

Is the above claim true if $G$ is replaced by a subgroup of $S_n$?

$\mathbf{For \: example},$ let $V=\mbox{Spec} \; k[x_1,y_1,x_2,y_2,x_3,y_3]$ take $\sigma = (12)(34)(56)$ to be in $S_6$ so that

$$\sigma\circ(x_1,y_1,x_2,y_2,x_3,y_3)=(y_1,x_1,y_2,x_2,y_3,x_3).$$ So $\sigma$ permutes $x_i$ to $y_i$ for each $i$.

Then writing $\left< \sigma\right>\leq S_6$ to be the subgroup of order $2$ generated by $\sigma$, $$k[V]^{\left< \sigma\right>} = k[x_1 + y_1, x_2 + y_2, x_3 + y_3, x_1 y_1, x_2 y_2, x_3 y_3, x_1 y_2 + x_2 y_1,$$

$$x_1 y_3+ x_3 y_1, x_2 y_3+x_3y_2, x_1x_2 x_3 + y_1 y_2 y_3].$$

Is there a relation among the generators of $k[V]^{\left< \sigma\right>}$?



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