Question

Let R = kG where k is any field and G is any group. Let I be the ideal of R generated by ab − ba for all a, b ∈ R. Show that R/I ∼ k[G/G'] as k-algebras, where G' denotes the commutator subgroup of G.

Answer 1

Let $G$ be the abelian group $G / G'$. Then $kG$ is a commutative ring, so the natural ring homomorphism $\varphi_0: R \to kG$ sends $ab - ba$ to zero, for all $a, b \in R$. This shows that $\varphi_0(I) = 0$, so $\varphi_0$ induces a $k$-algebra homomorphism $\varphi: R / I \to kG$. Next, we shall try to construct a $k$-algebra homomorphism $\psi: kG \to R / I$. Consider the group homomorphism $\psi_0: G \to \mathrm{U}(R / I)$ defined by $\psi_0(g) = g + I$. Since $\psi_0(gh) = gh + I = hg + I = \psi_0(hg)$ ($\forall g, h \in G$), $\psi_0$ induces a group homomorphism $G \to \mathrm{U}(R / I)$, which in turn induces a $k$-algebra homomorphism $\psi: kG \to R / I$. It is easy to check that $\psi$ and $\varphi$ are mutually inverse maps, so we have $R / I \sim kG$.