${I}=\Delta(G,G')$ is the smallest ideal of the group ring $\mathbb{Z}{G} $ such that $\mathbb{Z}{G}/{I}$ is a commutative ring

Question

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 $N\unlhd G$ , $\Delta(G,N)=\{\sum_{n\in N} \alpha_n(n-1) : \alpha_n \in \Bbb{Z}G\}$

You seem to be assuming we all know what $\Delta(G,G')$ is... — Mariano Suárez-Álvarez, Jul 23 '15 at 02:34
@MarianoSuárez-Alvarez It is basically defined as the kernel mentioned but I shall edit with another useful definition. — Bhaskar Vashishth, Jul 23 '15 at 02:34
The notation is used twice before defining it, and I had never seen it before. — Mariano Suárez-Álvarez, Jul 23 '15 at 02:58

score 0 · Answer 1 · answered Nov 30 '16 at 08:58

I suppose you have found it by yourself by now, but for completeness' sake I will write an answer here.

Let $R $ be any commutative ring. Let $J$ be an ideal of $ RG $. Then we can look at the images $\overline{g}$ of the group elements $g \in G $ in the quitient ring $ RG / J $. They should commute with eachother if the quotient ring is commutative. Hence, for any $g, h \in G $ it follows that $$ \overline {g^{-1}h^{-1} gh} = \overline {1}. $$ Or if we rewrite this, we find that $$ g^{-1}h^{-1} gh - 1 \in J. $$ As $J $ is an ideal, this shows that $$\Delta ( G, G') \subseteq J. $$ So any ideal whose quotient ring is commutative contains this partial augmentation ideal $\Delta ( G,G') $. As you stated yourself, $I:=\Delta (G,G') $ is an ideal. Moreover, $RG/I \cong R (G/G') $. Hence, I is the smallest ideal in $RG $ whose quotient ring is commutative.

${I}=\Delta(G,G')$ is the smallest ideal of the group ring $\mathbb{Z}{G} $ such that $\mathbb{Z}{G}/{I}$ is a commutative ring

1 Answers1