## 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

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

$=f(x)f(y).$

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. $\Box$