## The augmentation ideal of a group ring

Posted: October 27, 2011 in Group Algebras, Noncommutative Ring Theory Notes
Tags: , , ,

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. $\Box$

## Center of group algebras

Posted: February 8, 2011 in Group Algebras, Noncommutative Ring Theory Notes
Tags:

Theorem. Let $k$ be a field and let $G$ be a finite group. Let $\mathcal{C}_1, \cdots , \mathcal{C}_m$ be the conjugacy classes of $G.$ Let

$c_i = \sum_{g \in \mathcal{C}_i} g \in k[G], \ 1 \leq i \leq m.$

The set $\{c_1, \cdots , c_m \}$ is a $k$-basis for $Z(k[G]),$ the center of $k[G].$ Thus $\dim_k Z(k[G])$ is equal to the number of conjugacy classes of $G.$

Proof$c_1, \cdots , c_m$ are linearly independent over $k$ because if $i \neq j,$ then $\mathcal{C}_i \cap \mathcal{C}_j = \emptyset.$ So we only need to show that $\text{span} \{c_1, \cdots , c_m \} = Z(k[G]).$ Let $x = \sum_{g \in G} x_g g \in k[G],$  where $x_g \in k$ for all $g \in G.$

Claim. $x \in Z(k[G])$ if and only if $x_g = x_{h^{-1}gh}$ for all $g,h \in G.$

Proof  of the claim. Well, $x \in Z(k[G])$ if and only if $xh = hx$ for all $h \in G.$ Thus, $x \in Z(k[G])$ if and only if

$\sum_{g \in G} x_g g = \sum_{g \in G} x_g h g h^{-1}. \ \ \ \ \ \ \ \ (1)$

Clearly, fixing $h \in G,$ we have $\{hgh^{-1}: \ g \in G \} =G.$ So if we put $hgh^{-1}=u,$ then $g = h^{-1}uh$ and thus $(1)$ becomes

$\sum_{g \in G} x_g g = \sum_{g \in G} x_{h^{-1}gh} g. \ \ \ \ \ \ \ \ \ \ \ (2)$

The claim now follows from $(2).$

Now, if $1 \leq i \leq m,$ then, by definition, $g \in \mathcal{C}_i$ if and only if $h^{-1}gh \in \mathcal{C}_i$ for all $h \in G.$ So for every $g,h \in G,$ the coefficients of $g$ and $h^{-1}gh$ in $c_i$ are either both 1 or both 0. Thus, by the claim, $c_i \in Z(k[G])$ for all $i.$ So $\text{span} \{c_1, \cdots , c_m \} \subseteq Z(k[G]).$

Conversely, if $x = \sum_{g \in G} x_g g \in Z(k[G]),$ then, by the claim, $x_g = x_{g'}$ for all $g$ and $g'$ which are in the same conjugacy class. Thus if for every $i$ we choose a $g_i \in \mathcal{C}_i,$ then

$x = \sum_{i=1}^m x_{g_i} c_i.$

So $Z(k[G]) \subseteq \text{span} \{c_1, \cdots , c_m \}. \ \Box$

## Semiprimitivity of k[G] for finite groups G

Posted: January 9, 2011 in Group Algebras, Noncommutative Ring Theory Notes
Tags: , , ,

Theorem (Maschke, 1899) Let $k$ be a field and let $G$ be a finite group of order $n.$ Let $R:=k[G].$ Then $R$ is semiprimitive if and only if $char(k) \nmid n.$

Proof. Let $G = \{g_1, \cdots , g_n \}$ where $g_1=1.$ Suppose first that $char(k) \nmid n$ and consider the algebra homomorphism $\rho : R \longrightarrow End_k(R)$ defined by $\rho(r)(x)=rx$ for all $r,x \in R.$ Define $\alpha : R \longrightarrow k$ by $\alpha(r) = Tr(\rho(r))$ for all $r \in R.$ Note that here $Tr(\rho(r))$ is the trace of the matrix corresponding to the linear transformation $\rho(r)$ with respect to the ordered basis $\{g_1, \cdots , g_n \}.$ We remark a few points about $\alpha$:

1) $\alpha(1)=n$ because $\rho(1)$ is the identity map of $R.$

2) If $1 \neq g \in G,$ then $\alpha(g)=0.$ The reason is that $\rho(g)(g_i)=gg_i \neq g_i$ for all $i$ and thus the diagonal entries of the matrix of $\rho(g)$ are all zero and so $\alpha(g)=Tr(\rho(g))=0.$

3) If $r$ is a nilpotent element of $R,$ then $\alpha(r)=0$ because then $r^m=0$ for some $m$ and thus $(\rho(r))^m = \rho(r^m)=0.$ So $\rho(r)$ is nilpotent and we know that the trace of a nilpotent matrix is zero.

Now let $c \in J(R).$ Since $R$ is finite dimensional over $k,$ it is Artinian and hence $J(R)$ is nilpotent. Thus $c$ is nilpotent and therefore, by 3), $\alpha(c)=0.$ Let $c = \sum_{i=1}^n c_i g_i,$ where $c_i \in k.$ Then

$0=\alpha(c)=\sum_{i=1}^n c_i \alpha(g_i)=c_1n,$

by 1) and 2). It follows that $c_1=0$ because $char(k) \nmid n.$ So the coefficient of $g_1=1$ of every element in $J(R)$ is zero. But for every $i,$ the coefficient of $g_1=1$ of the element $g_i^{-1}c \in J(R)$ is $c_i$ and so we must have $c_i=0.$ Hence $c = 0$ and so $J(R)=\{0\}.$

Conversely, suppose that $char(k) \mid n$ and put $x = \sum_{i=1}^n g_i.$ Clearly $xg_j = x$ for all $j$ and hence

$x^2=x\sum_{j=1}^n g_j = \sum_{j=1}^n xg_j=\sum_{j=1}^n x = nx = 0.$

Thus $kx$ is a nilpotent ideal of $R$ and so $kx \subseteq J(R).$ Therefore $J(R) \neq \{0\}$ because $kx \neq \{0\}. \ \Box$

## The group ring isomorphism problem

Posted: June 6, 2010 in Group Algebras, Noncommutative Ring Theory Notes
Tags: , ,

Let $k$ be a field or $\mathbb{Z}$ and let $G_1,G_2$ be two groups. It is clear that if $G_1 \cong G_2,$ then $k[G_1] \cong k[G_2],$ as $k$ algebras. The group ring isomorphism problem is this question that whether or not $k[G_1] \cong k[G_2],$ as $k$ algebras, implies $G_1 \cong G_2$. Another version of the group ring isomorphism problem is this: given a group $G_1$ and a field $k$ find all groups $G_2$ such that $k[G_1] \cong k[G_2],$ as $k$ algebras. These questions have been answered in special cases only. For example, an old result due to Perlis and Walker states that if $G_1,G_2$ are finite abelian groups and $\mathbb{Q}[G_1] \cong \mathbb{Q}[G_2],$ as $\mathbb{Q}$ algebras, then $G_1 \cong G_2.$ In 2001 Hertweek found two non-isomorphic groups $G_1, G_2$ of order $2^{21}97^{28}$ such that $\mathbb{Z}[G_1] \cong \mathbb{Z}[G_2].$

We now show that it is quite possible to have $\mathbb{C}[G_1] \cong \mathbb{C}[G_2]$ and $G_1 \ncong G_2$:

Theorem. If $G_1, G_2$ are two finite abelian groups of order $n,$ then $\mathbb{C}[G_1] \cong \mathbb{C}[G_2] \cong \mathbb{C}^n$ as $\mathbb{C}$-algebras.

Proof. We have already seen in this post that $J(\mathbb{C}[G])=(0)$ for any group $G.$ So if $G$ is a finite abelian group of order $n,$ then $\mathbb{C}[G]$ is a commutative semisimple algebra. Thus, since $\mathbb{C}$ is algebraically closed, the Arrin-Wedderburn theorem gives $\mathbb{C}[G] \cong \mathbb{C}^n$ as $\mathbb{C}$-algebras. $\Box$

Example. Let $G_1$ be the Klein four-group and let $G_2$ be the cyclic group of order four. Then $G_1 \ncong G_2$ but, by the theorem, $\mathbb{C}[G_1] \cong \mathbb{C}[G_2] \cong \mathbb{C}^4$ as $\mathbb{C}$-algebras.

## Semiprimitivity of C[G]

Posted: December 2, 2009 in Group Algebras, Noncommutative Ring Theory Notes
Tags: , , , ,

Notation. For a ring $R$ let $J(R)$ be the Jacobson radical of $R.$

Definition. Recall that if $k$ is a field and $G$ is a group, then the group algebra $k[G]$ has two structures. Firstly, as a vector space over $k,$ it has $G$ as a basis, i.e. every element of $k[G]$ is uniquely written as $\sum_{g \in G} x_g g,$ where $x_g \in k.$ In particular, $\dim_k k[G]=|G|,$ as cardinal numbers. Secondly, multiplication is also defined in $k[G].$ If $x = \sum_{g \in G} x_g g$ and $y = \sum_{g \in G} y_g g$ are two elements of $k[G],$ then we just multiply $xy$ in the ordinary fashion using distribution law. To be more precise, we define $xy = \sum_{g \in G} z_g g,$ where $z_g = \sum_{rs=g} x_r y_s.$

We are going to prove that $J(\mathbb{C}[G])=0,$ for every group $G.$

Lemma. $J(\mathbb{C}[G])$ is nil, i.e. every element of $J(\mathbb{C}[G])$ is nilpotent.

Proof. If $G$ is countable, we are done by this theorem. For the general case, let $\alpha \in J(\mathbb{C}[G]).$ So $\alpha =\sum_{i=1}^n c_ig_i,$ for some $c_i \in \mathbb{C}, \ g_i \in G.$ Let $H=\langle g_1,g_2, \cdots , g_n \rangle.$ Clearly $\alpha \in H$ and $H$ is countable. So to complete the proof, we only need to show that $\alpha \in J(\mathbb{C}[H]).$ Write $G = \bigcup_i x_iH,$ where $x_iH$ are the distinct cosets of $H$ in $G.$ Then $\mathbb{C}[G]=\bigoplus_i x_i \mathbb{C}[H],$ which means $\mathbb{C}[G]=\mathbb{C}[H] \oplus K,$ for some right $\mathbb{C}[H]$ module $K.$ Now let $\beta \in \mathbb{C}[H].$ Since  $\alpha \in J(\mathbb{C}[G]),$ there exists $\gamma \in \mathbb{C}[G]$ such that $\gamma (1 - \beta \alpha ) = 1.$ We also have $\gamma = \gamma_1 + \gamma_2,$ for some $\gamma_1 \in \mathbb{C}[H], \ \gamma_2 \in K.$ That now gives us $\gamma_1(1 - \beta \alpha)=1. \ \Box$

Theorem. $J(\mathbb{C}[G])=0,$ for any group $G.$

Proof. For any $x =\sum_{i=1}^n c_i g_i\in \mathbb{C}[G]$ define

$x^* = \sum_{i=1}^n \overline{c_i} g_i^{-1}.$

It’s easy to see that $xx^*=0$ if and only if $x=0$ and for all $x,y \in \mathbb{C}[G]: \ (xy)^*=y^*x^*.$ Now suppose that $J(\mathbb{C}[G]) \neq 0$ and let $0 \neq \alpha \in J(\mathbb{C}[G]).$ Put $\beta = \alpha \alpha^* \in J(\mathbb{C}[G]).$ By what I just mentioned $\beta \neq 0$ and $(\beta^m)^* = (\beta^*)^m=\beta^m,$ for all positive integers $m.$ By the lemma, there exists $k \geq 2$ such that $\beta^k = 0$ and $\beta^{k-1} \neq 0.$ Thus $\beta^{k-1} (\beta^{k-1})^* = \beta^{2k-2} = 0,$ which implies that $\beta^{k-1} = 0.$ Contradiction! $\Box$

Corollary. If $G$ is finite, then $\mathbb{C}[G]$ is semisimple.

Proof. We just proved that $J(\mathbb{C}[G])=(0).$ So we just need to show that $\mathbb{C}[G]$ is Artinian. Let

$I_1 \supset I_2 \supset \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$

be a descending chain of left ideals of $\mathbb{C}[G].$ Obviously each $I_j$ is a $\mathbb{C}$-subspace of $\mathbb{C}[G].$ Thus each $I_j$ is finite dimensional because $\dim_{\mathbb{C}} \mathbb{C}[G]=|G| < \infty.$ Hence $(*)$ will stablize at some point because $\dim_{\mathbb{C}} I_1 < \infty$ and $\dim_{\mathbb{C}}I_1 > \dim_{\mathbb{C}} I_2 > \cdots .$ Thus $\mathbb{C}[G]$ is (left) Artinian and the proof is complete because we know a ring is semisimple if and only if it is (left) Artinian and its Jacobson radical is zero. $\Box$