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

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