Is the ring of polynomial invariants of a finite perfect group an UFD? Let $G$ be a finite group. $G$ acts on $\mathbb K[x_1,...,x_n]$ by automorphisms fixing $K$. $\mathbb K[x_1,...,x_n]^G=\{ T\in \mathbb K[x_1,...,x_n],\forall \sigma \in G, T^{\sigma}=T\}$ is the ring of invariants. 

Is it true that $\mathbb K[x_1,...,x_n]^G$ is a unique factorization domain if commutator subgroup of $G$ equals to $G$?


Let $G = S_n$ act on the polynomial ring in the standard way, by permutation of the generators. Then $K[x_1, \cdots, x_n]^G \cong k[s_1, \cdots, s_n]$ is a standard result on symmetric polynomials, and $k[s_1, \cdots, s_n]$ is a UFD. However, $[G:G] = [S_n:S_n] = A_n \neq G$.
I don't know offhand the answer to your question regarding the converse.
