Archive for the ‘Group Algebras’ Category

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

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Notation. 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.

Theorem. 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.

Proofc_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). \ \Box

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

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 

Let k be a field or \mathbb{Z} and G_1,G_2 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].

For now, we’ll only show that it is possible to have k[G_1] \cong k[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 Wedderburn-Artin theorem gives us \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.

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