2 Group Algebra

group algebra

If the vector space $V$ over $F$ has a basis $\{g|g\in G\}$ and $\dim V=|G|$, we have a special $FG$-module: group algebra, which corresponds the “regular representation”.

The group algebra is denoted by $FG$, and $\{g|g\in G\}$ is called the natural basis of $FG.$

Maschke’s theorem

This theorem shows that $FG$ is completely reducible if $F=\mathbb{C}$ or $\mathbb{R}$.

Maschke's theorem

Let $G$ be a finite group, let $F$ be $\mathbb{R}$ or $\mathbb{C}$, and let $V$ be an FG-module. If $U$ is an FG-submodule of $V$, then there is an FG-submodule $W$ of $V$ such that$$ V=U\oplus W.

Remark. The Maschke’s theorem may fail, if one of the following holds.

• $G$ is an infinite group.
• The characteristic of $F$ is not $0$. See Maschke’s theorem for more details.

Schur’s lemma

Theorem

Let $V$ and $W$ be irreducible $\mathbb{C}G$-modules.

• If $\theta :V\to W$ is a $\mathbb{C}G$-homomorphism, then either $\theta$ is a CG-isomorphism, or $v\theta =0$ for all $v\in V$.
• If $\theta :V\to V$ is a $\mathbb{C}G$-isomorphism, then $\theta$ is a scalar multiple of the identity endomorphism $1_V$.

\begin{proof} Since $V$ is irreducible and $\ker \theta ⩽V$ is an submodule, we have that $\ker \theta =0\text{mbox}orV$. If $\ker \theta =V$, then $v\theta =0$ for all $v\in V$. Otherwise, when $\ker \theta =0$, $\theta$ is an isomorphism.

Now suppose that $V=W$. As $\theta$ is a linear transformation, it has at least one eigenvalue $\lambda$. Then $\theta -\lambda I$ is also a $\mathbb{C}G$-homomorphism and $\ker (\theta -\lambda I)=V$. It follows that $v\theta =\lambda v$. \end{proof}

Remark. Note that $V$ should be a finite-dimensional vector space, otherwise the eigenvalue of $\theta$ may not exist.

Applications of Schur’s lemma

Proposition

Let $V$ be a FG-module. If every $FG$-homomorphism $\theta :V\to V$ is $\theta =\lambda I$, then $V$ is irreducible.

Corollary

Let $\rho :G\to \mathrm{GL}(n,\mathbb{C})$ be a representation of G. Then $\rho$ is irreducible if and only if every matrix A which satisfies $A(g\rho )=(g\rho )A$ for all $g\in G$ has the form $A=\lambda I_n$ with $\lambda \in \mathbb{C}$.

Structure of Group Algebra $FG$

Warning

These conclusions hold only for $\mathbb{C}G$-module, as all of them need Maschke’s theorem and Schur’s lemma.

By Maschke’s theorem, we know that $FG=U_1\oplus ...\oplus U_r$ when $F=\mathbb{R}$ or $\mathbb{C}$, where $U_i$ are irreducible FG-submodules.

Abelian Group

Let $G$ be an abelian group and $V$ be the group algebra $FG$. Note that for any $x\in G$, ${\theta }_x:v\mapsto vx$ is an $FG$-homomorphism, as $vxg=vgx$ for all $v\in V$ and $g\in G$.

For any irreducible submodule $W\subseteq V$, there exists a ${\lambda }_W$ such that ${\theta }_x{|}_W:v\mapsto {\lambda }_{W}v$ by Schur’s lemma. It follows that $W=⟨v⟩$ is a submodule of $V$. Therefore, each irreducible submodule of $FG$ is of dimension $1$. In the other word, suppose $\rho$ is the regular representation, then there exists a basis such that $\rho (x)$ is a diagonal matrix, as the following proposition shows.

Proposition

Let $G$ be a finite group, and let $V$ be a $\mathbb{C}G$-module. Then there is a basis $\mathcal{B}$ such that $[g{]}_{\mathcal{B}}$ is diagonal and the entries on the diagonal of $[g{]}_{\mathcal{B}}$ are nth roots of unity, where n is the order of $g$.

Remark. This proposition shows that every representation of a element of a finite group is diagonalizable. We can use it to prove that ”$A$ is diagonalizable if $A^n=I$ for some $n$“.

By the fundamental theorem of abelian group, the abelian group $G$ can be written as $G=C_{n_1}\times ...\times C_{n_r}$. Then there are $n$ irreducible $\mathbb{C}G$-modules, whose corresponding representation ${\rho }_{{\lambda }_1\cdots {\lambda }_r}$ is defined as $g_i\rho =({\lambda }_i)$ for any generator $g_i$ of $C_{n_i}$, where

$$n=|\{\lambda ={\Pi }_{i=1}^r{\lambda }_i:{\lambda }_i^{n_i}=1\}|=|G|.$$

So we get the following theorem.

Theorem

Let $G$ be the abelian group $C_{n_1}\times \cdots \times C_{n_r}$. The representations ${\rho }_{{\lambda }_1\cdots {\lambda }_r}$ of $G$ constructed above are irreducible and have degree $1$. There are $|G|$ of these representations, and every irreducible representation of $G$ over $\mathbb{C}$ is equivalent to precisely one of them.

Furthermore, the converse argument is also true.

Proposition

Suppose that $G$ is a finite group such that every irreducible $\mathbb{C}G$-module has dimension 1. Then $G$ is abelian.

Non-abelian Group

By Maschke’s theorem, We know that $FG=U_1\oplus ...\oplus U_s$ where $U_i$ are irreducible $FG$-submodule.

Now consider the number of $j$ such that $U_i\cong U_j$ for a fixed $U_i$, and denote the number as $d_i$. By Schur’s lemma, there is $d_i=\dim ({\mathrm{Hom}}_{\mathbb{C}G}(U_i,\mathbb{C}G))$, because

• $\dim ({\mathrm{Hom}}_{\mathbb{C}G}(U_i,U_j))=1$ if $U_i\cong U_j$, and
• $\dim ({\mathrm{Hom}}_{\mathbb{C}G}(U_i,U_j))=0$ if $U_i\ncong U_j$.

Furthermore, there is

$$\begin{array}{rl}\dim ({\mathrm{Hom}}_{\mathbb{C}G}(U_i,\mathbb{C}G))& =\sum _{j=1}^s\dim ({\mathrm{Hom}}_{\mathbb{C}G}(U_i,U_j))\\ & =\sum _{j=1}^s\dim ({\mathrm{Hom}}_{\mathbb{C}G}(U_j,U_i))\\ & =\dim ({\mathrm{Hom}}_{\mathbb{C}G}(\mathbb{C}G,U_i))\end{array}$$

and so we have that $d_i=\dim ({\mathrm{Hom}}_{\mathbb{C}G}(\mathbb{C}G,U_i))$.

Suppose that $u_1,\cdots ,u_d$ is a set of basis of $U_i$. Define ${\phi }_i:\mathbb{C}G\to U_i$ by $r\mapsto u_{i}r$. For any $\phi \in {\mathrm{Hom}}_{\mathbb{C}G}(\mathbb{C}G,U_i)$, $1\phi ={\lambda }_1{u}_1+\cdots +{\lambda }_d{u}_d$, then $r\phi =(1\phi )r=r({\lambda }_1{u}_1+\cdots +{\lambda }_d{u}_d)$ and so $\phi ={\lambda }_1{\phi }_1+\cdots +{\lambda }_d{\phi }_d$. Then we have

$$\dim ({\mathrm{Hom}}_{\mathbb{C}G}(\mathbb{C}G,U_i))=\mathrm{span}({\phi }_1,...,{\phi }_s)=\dim U_i,$$

and get the following theorem.

Theorem

Suppose that $\mathbb{C}G=U_1\oplus \cdots \oplus U_r$ a direct sum of irreducible $\mathbb{C}G$-submodule. If $U$ is any irreducible $\mathbb{C}G$-module, then the number of $\mathbb{C}G$-module $U_i$ with $U_i\cong U$ is equal to $\dim U$.

In particular, let $V_1,\cdots ,V_k$ form a complete set of non-isomorphic irreducible $\mathbb{C}G$-modules. Then ${\sum }_{i=1}^k(\dim V_i{)}^2=|G|$.

Center of $FG$

Define the center of $\mathbb{C}G$ as

$$Z(\mathbb{C}G)=\{z\in \mathbb{C}G:zr=rz,\forall r\in \mathbb{C}G\}.$$

Each element of $Z(\mathbb{C}G)$ is a sum of elements of some conjugate classes. Notice that $Z(\mathbb{C}G)$ is a subspace of $\mathbb{C}G$, so $\dim Z(\mathbb{C}G)=\#$conjugate classes of $G$.

For any $r\in Z(\mathbb{C}G)$, $vrg=vgr$ and $\theta :v\to vr$ is a $\mathbb{C}G$-homomorphism. If $\theta :V\to V$ for an irreducible $\mathbb{C}G$-submodule, then $vr=\lambda v$. Similarly, for an element $z\in Z(G)$, $\theta :v\to vz$ is also a $\mathbb{C}G$-homomorphism. Then we get a proposition as follows.

Proposition

If there exists a faithful irreducible $\mathbb{C}G$-module $V$, then $Z(G)$ is cyclic.

\begin{proof} For any $z\in Z(G)$,

v\mapsto vz$$ is a $\mathbb CG$-homomorphism and $vz=\lambda_zv$ by [[Schur Lemma|Schur's lemma]]. Since $V$ is faithful, we have that $\lambda_x\neq\lambda_y$ if $x\neq y$. Otherwise, there exists $1\neq xy^{-1}\in Z(G)$ such that $v(xy^{-1})=\lambda_x\lambda_y^{-1}v=v$ for all $v\in V$. Thus, $Z(G)$ is a subgroup of $\mathbb C^{*}$ and so it is cyclic. `\end{proof}`