## Semiprimitivity of C[G]

Posted: December 2, 2009 in Group Algebras, Noncommutative Ring Theory Notes
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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. $\Box$