How can I prove that if ${G}$ is a group then ${I}=\Delta(G,G')$ is the smallest ideal of the group ring $\mathbb{Z}{G} $ such that $\mathbb{Z}{G}/{I}$ is a commutative ring?
It is an ideal is clear, as it is the kernel of homomorphism ${p}:\Bbb{Z}{G}\to \Bbb{Z}({G}/{G'})$ where $\sum_{{g\in G}} {a}_{g}{g}\to \sum_{g\in G} {a}_{g}\bar{{g}}$ where $\bar\ :{G\to G/G'}$ is the canonical homomorphism.
Also as $\frac{\Bbb{Z}{G}}{\Delta(G,G')}\cong \Bbb{Z}({G}/{G'})$ and ${G}/{G'}$ is abelian implies $\mathbb{Z}{G}/{I}$ is commutative. But I am having trouble showing that it is smallest such ideal. Let ${J}$ be any other such ideal such that $\mathbb{Z}{G}/{J}$ is commutative then I need to show that ${I\subset J}$.
My question is that it doesnot always have to that $\frac{\Bbb{Z}{G}}{{J}}\cong\Bbb{Z}({G/N})$ for some ${N}\unlhd {G}$ so that if $\Bbb{Z}({G/N})$ is commutative I can say ${G'}\subset N$ which gives a monomorphism ${G/N}\to {G/G'}$and thus $\mathbb{Z}{G}/{J}\subset \mathbb{Z}{G}/{I}$ which would imply that ${I}\subset {J}$. Am I missing something here?
Definition- For $N\unlhd G$ , $\Delta(G,N)=\{\sum_{n\in N} \alpha_n(n-1) : \alpha_n \in \Bbb{Z}G\}$