The augmentation ideal of a group ring

Posted: October 27, 2011 in Group Algebras, Noncommutative Ring Theory Notes
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Lemma. Let R be a commutative ring and let G be a group. We define the map f : R[G] \longrightarrow R by

f(\sum_{g \in G} r_g g)=\sum_{g \in G} r_g.

Then f is an onto ring homomorphism and \{g - 1_G: \ g \in G, g \neq 1_G \} is a basis for the free R-module \ker f.

Proof. Obviously f is well-defined, additive and onto. Now, let x = \sum_{g \in G} r_g g, \ y = \sum_{g \in G} s_g g. Then

f(xy) = f(\sum_{g \in G} (\sum_{g_1g_2=g} r_{g_1}s_{g_2}) g)=\sum_{g \in G} \sum_{g_1g_2=g} r_{g_1}s_{g_2}=(\sum_{g \in G} r_g)(\sum_{g \in G} s_g)


So f is a ring homomorphism. Therefore \ker f is an ideal of R[G] and hence an R-module. Now, x = \sum_{g \in G} r_g g \in \ker f if and only if \sum_{g \in G}r_ g = 0 if and only if

x = x - (\sum_{g \in G}r_g)1_G=\sum_{g \in G}r_g(g-1_G).

Thus \ker f, as an R-module, is generated by the set \{g - 1_G : \ g \in G \}. Then, obviously, the set B=\{g-1_G: \ g \in G, g \neq 1_G\} still generates \ker f. To show that B is a basis for \ker f, as an R-module, we suppose that \sum_{g \in B}r_g (g-1_G)=0. Then

\sum_{g \in B}r_g g = (\sum_{g \in B}r_g)1_G.

But g \neq 1_G for all g \in B, and so r_g = 0 for all g \in B. \ \Box

Definition. The ring homomorphism f, as defined in the lemma, is called the augmentation map and \ker f is called the augmentation ideal of R[G].

Hurewicz Theorem. Let G be a group with the commutator subgroup G'. Let I be the augmentation ideal of \mathbb{Z}[G] and consider I as an additive group. Then \displaystyle \frac{G}{G'} \cong \frac{I}{I^2}.

Proof. Define the map \displaystyle \varphi : G \longrightarrow \frac{I}{I^2} by \varphi(g)=g-1_G + I^2. Clearly \varphi is well-defined because g - 1_G \in I for all g \in G. Also, since (g_1 - 1_G)(g_2-1_G) \in I^2, we have

\varphi(g_1g_2)=g_1g_2-1_G + I^2 = g_1g_2 -1_G - (g_1-1_G)(g_2-1_G) + I^2=

g_1 - 1_G + g_2 - 1_G + I^2 = \varphi(g_1) + \varphi(g_2).

Thus \varphi is a group homomorphism. So \varphi(g^{-1})=-\varphi(g) and therefore, since I is an abelian group, we have \varphi(g_1g_2g_1^{-1}g_2^{-1})=\varphi(g_1) + \varphi(g_1^{-1})+\varphi(g_2)+\varphi(g_2^{-1})=0. Thus G' \subseteq \ker \varphi and hence we have a group homomorphism \displaystyle \overline{\varphi}: \frac{G}{G'} \longrightarrow \frac{I}{I^2} defined by

\overline{\varphi}(gG')=g - 1_G +I^2.

Now, to show that \overline{\varphi} is an isomorphism, we will find an inverse for it. Define the map \psi : I \longrightarrow G/G' by \psi(\sum_{g \in G} n_g(g-1_G))=(\prod_{g \in G} g^{n_g})G'. Note that since g_1g_2G'=g_2g_1G' for all g_1,g_2 \in G, the map \psi is a well-defined group homomorphism. Now,

\psi((g_1 - 1_G)(g_2-1_G))=\psi(g_1g_2 -1_G -(g_1-1_G)-(g_2-1_G))=g_1g_2g_1^{-1}g_2^{-1}G' =G'.

So I^2 \subseteq \ker \psi because, by the lemma, the set \{(g_1-1_G)(g_2-1_G): \ g_1,g_2 \in G\} generates the additive group I^2. Hence the map \displaystyle \overline{\psi} : \frac{I}{I^2} \longrightarrow \frac{G}{G'} defined by

\overline{\psi}(\sum_{g \in G}n_g (g - 1_G)+ I^2) = (\prod_{g \in G} g^{n_g})G'

is a well-defined group homomorphism. It is now easy to see that both \overline{\varphi} \circ \overline{\psi} and \overline{\psi} \circ \overline{\varphi} are identity maps. \Box

  1. Hamed says:

    Thank you very much…

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