Representation of Finite Groups

Definition 8.1   Let $G$ be a finite group. A representation $\rho$ of $G$ is is a homomorphism of $G$ into the group $\textrm{GL}(V)$ of invertible linear transformations of a finite dimensional vector space $V$ over a field $F$. The dimension of $V$ is called the degree of the representation.

Some easy facts concerning representations:

1. $\rho(g_{1}g_{2})=\rho(g_{1})\rho(g_{2})$. Particularly, $\rho(g^{m})=\rho(g)^{M}$.
2. $\rho(g^{-1})=\rho(g)^{-1}$.
3. $\rho(1)=\rho(1)$.
4. By choosing a base for $V$, we can write $\rho$ in a matrix form, and get an homomorphism of $G$ into the group $\textrm{GL}_{n}(F)$.

Do we have two unrelated definitions of a ``representation''? No. If $G$ is a finite group and $F$ is some field, then define the set $F[G]$ to be the a vector space over $F$ having the set $G$ has a base. The set can be made into a ring by defining:

$$(\sum_{g\in G}\alpha_{g}g)(\sum_{h\in G}\beta_{h}h)=\sum_{g,h}\alpha_{g}\beta_{h}gh$$

for $\alpha_{g},\beta_{h}\in F$.

If $\rho$ is a group representation of $G$ acting of $V$, then we can extend $\rho$ to $F[G]$, by defining:

$$\rho'(\sum\alpha_{g}g)=\sum_{g\in G}\alpha_{g}\rho(g)$$

It follows that $\rho'$ is a homomorphism of the ring $F[G]$ into $\textrm{End}_{F}V$. Conversely, if $\rho'$ is a homomorphism of the ring $F[G]$ into $\textrm{End}_{F}V$ where $V$ is finite dimensional vector space over $F$, then the restriction of $\rho'$ to $G$ is a representation, since $\rho'(1)=1$ and thus $\rho'(g)\in\textrm{GL}(V)$. We've seen that each representation of a ring $F[G]$ can make it a module:

$$(\sum\alpha_{g}g)x=\sum\alpha_{g}\rho(g)x$$

for $x\in V$. This shows that representations of a group $G$ are equivalent to $F[G]$-modules, which as $F$-modules, are finite dimensional.

Definition 8.2   Let $\rho_{1}$ and $\rho_{2}$ be representations of a group $G$. We say that $\rho_{1}$ and $\rho_{2}$ are equivalent if the corresponding $F[G]$-modules $V_{1}$ and $V_{2}$ are isomorphic.

Let $\rho_{1}$ and $\rho_{2}$ be equivalent representations of a group $G$. Denote by $\mu:V_{1}\rightarrow V_{2}$ the isomorphism of $V_{1}$ onto $V_{2}$ as $F[G]$-submodules. By definition we have for every $g\in G$ and $v\in V_{1}$ that:

$$\mu(gv)=g\mu(v)$$

which is equivalent to:

$$\mu(\rho_{1}(g)v)=\rho_{2}(g)\mu(v)$$

which implies that:

$$\rho_{1}(g)=\mu^{-1}\rho_{2}(g)\mu$$

Theorem 8.3   $\mu$ is just the module

Let $\rho$ be a representation of $G$ acting on $V$ and let $U$ be a submodule of $V$ as $F[G]$-module. Thus, $U$ is a subspace of $V$ which is $\rho(G)$ invariant: it is stabilized by every $\rho(g)$ where $g\in G$. Thus, in restricting $\rho(g)$ to $U$, we can get a new representation $\rho\vert U$. We shall call this a sub-representation of $\rho$. If $V=U\oplus U'$ where $U$ and $U'$ are submodules, then we can write $\rho=(\rho\vert U)\oplus(\rho\vert U')$ and say that $\rho$ is the direct sum of the sub-representations. Let $p$ be the projection of $V$ on $U$ determined by the above decomposition. Given any $x\in V$, write $x=y+y'$ where $y\in U$ and $y'\in U'$ then we have, for every $g\in G$:

$$\rho(g)x=\rho(g)y+\rho(g)y'$$

since $U$ and $U'$ are invariant subspaces of $\rho(g)$, we have that $\rho(g)y\in U$ and $\rho(g)y'\in U'$ it follows that

$$p\rho(g)y=\rho(g)y=\rho(g)py$$

and

$$p\rho(g)y'=0=\rho(g)py'$$

therefore,

$$p\rho(g)x=p\rho(g)(y+y')=\rho(g)p(y+y')=\rho(g)px$$

Thus, $p$ commutes with every $\rho(g)$. Conversely, suppose $U$ is a $\rho(G)$-invariant subspace of $V$, and there exists a projection $\rho$ of $V$ on $U$ that commutes with every $\rho(g)$. Write $V=pV\oplus(1-p)V$ (as subspaces, not submodules yet). Also,

$$\rho(g)(1-p)V=(\rho(g)-\rho(g)p)V=(1-p)\rho(g)V=(1-p)V$$

The last one holds since $\rho(g)$ is 1-1. This shows that $U'=(1-p)V$ is a $\rho(G)$ invariant subspace. Denote $U=pV$, then $V=U\oplus U'$.

We call a representation $\rho$of $G$ irreducible (or completely irreducible) if the corresponding $F[G]$-module is irreducible (completely irreducible). We have the following theorem:

Theorem 8.4 (Maschke's Theorem)   Every representation $\rho$ of a finite group $G$ acting on a vector space $V$ over a field $F$ such that $\textrm{char}F\nmid\vert G\vert$ is completely reducible.

Proof. Let $U$ be a submodule of $V$. We need to show that there is submodule $U'$ of $V$ such that $V=U\oplus U'$. Let $p_{0}$ be any projection of $V$ on $U$. Define:

$$p=\frac{1}{\vert G\vert}\sum_{g\in G}\rho(g)^{-1}p_{0}\rho(g)$$

Clearly, $p\in\textrm{End}_{F}V$. For any $x\in V$ we have $p_{0}\rho(g)x\in U$, since $p_{0}$ is a projection. Since $U$ is $\rho(G)$-invariant subspace, then $\rho(g^{-1})p_{0}\rho(g)x\in U$ and thus $pV\subset U$. If $y\in U$, then $\rho(g)y\in U$. Hence, $p_{0}\rho(g)y=\rho(g)y$, which implies that $\rho(g^{-1})p_{0}\rho(g)y=y$ and therefore $py=y$ for $y\in U$. This shows that $p$ is a projection on $U$. For any $g\in G$, consider:

$$\rho(h^{-1})p\rho(h) = \frac{1}{\vert G\vert}\sum_{g\in G}\rho(h^{-1}g^{-1})p_{0}\rho(gh)$$

$$= \frac{1}{\vert G\vert}\sum_{g\in G}\rho(gh)^{-1}p_{0}\rho(gh{})$$

$$= \frac{1}{\vert G\vert}\sum_{g\in G}\rho(g)^{-1}p_{0}\rho(g)$$

$$= p$$

and thus $p$ commutes with every $\rho(g)$. Then $V=U\oplus U'$ where $U'=(1-p)V$ and since $p$ commutes with every $\rho(g)$ we have that $U'$ is $\rho(G)$-invariant and thus submodule. $\qedsymbol$

From now on, we assume that $F=\mathbb{C}$, the field of complex numbers. A representation $\rho$ of $G$ acting on a $V$ where $V$ is a finite dimensional vector space over the field $\mathbb{C}$ is called a complex representation. It follows from the above that every complex representation is completely reducible. Thus, $V$ decomposes as $V=V_{1}^{(1)}\oplus\cdots\oplus V_{1}^{(m_{i})}\oplus V_{2}^{(1)}\oplus\cdots\... ...V_{2}^{(m_{2})}\oplus\cdots\oplus V_{r}^{(1)}\oplus\cdots\oplus V_{r}^{(m_{r})}$ where each $V_{i}^{(k)}$ are irreducible and $V_{i}^{(k)}\cong V_{j}^{(l)}$ if and only if $i=j$. If $\rho_{i}$ is any irreducible representation of $G$ equivalent to the sub-representation determined by $V_{i}^{(k)}$, then we write:

$$\rho\sim m_{1}\rho_{1}\oplus m_{2}\rho_{2}\oplus\cdots\oplus m_{r}\rho_{r}$$