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 $G$?

 A: Let $P$ be an irreducible polynomial. Any element $g$ of $G$ maps $P$ to some irreducible polynomial, because action is invertible. This polynomial $gP$ may be either proportional to $P$ or coprime to $P$. Let $Q = P \cdot (g_1 P) \cdot \ldots \cdot (g_k P)$ be a product of polynomials in the "essential orbit" of $P$: all pairwise non-proportional irreducible polynomials in the orbit. The action of $g \in G$ on $Q$ permutes the factors and multiplies them by some field elements, so $gQ = \phi(g) Q$, where $\phi\colon G \to k^*$ is some function on $G$. It is easy to check that $\phi$ is a homomorphism. As $k^*$ is commutative, the commutant of $G$ belongs to the kernel of $\phi$. Thus if $G$ is perfect, $Q$ is necessarily invariant.
Now it's straightforward to see that invariant polynomials over such $G$ form an UFD. Indeed, take an invariant polynomial $R\in k[x_1, \ldots, x_n]^G$. Let $P$ be any irreducible divisor of $R$ (not invariant, just a polynomial). As $R$ is invariant, it obviously is divisible by $Q$, but $Q$ is invariant, hence the quotient is invariant too. Any decomposition of $R$ into irreducible invariant polynomials contains a multiple of $P$ (after all, polynomial ring itself is an UFD), hence one of the factors is $Q$.
A: 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.
