2 Maschke Theorem

Let $G$ be a group, $k$ a field and $V$ a vector space over $k$. Assume that $V$ is a $G$-module and $\rho :G\to \mathrm{GL}(V)$ is the corresponding representation.

Maschke theorem

Let $G$ be a finite group and $\mathrm{char}k\nmid |G|$. Let $V$ be a $G$-module and $W$ a $G$-submodule. Then there exists a $G$-submodule $W^{\prime }$ such that $V=W\oplus W^{\prime }$.

\begin{proof} Consider the subspace $W\subset V$ and $V$ can be decomposed as $V=W\oplus W_0$. We have surjective map $\pi :V\to W,v\to w$. Define

$${\pi }_0:V\to W,v\mapsto \frac{1}{|G|}\sum _{g\in G}g\pi (g^{-1}v).$$

Then we have ${\pi }_0(w)=\frac{1}{|G|}{\sum }_{g\in G}g(g^{-1}w)=w$ and so ${\pi }_0$ is surjective. Therefore, $V=W\oplus \ker {\pi }_0$ as vector space. We now prove that $\ker {\pi }_0$ is a $G$-submodule. For any $h\in G$ and $v\in \ker {\pi }_0$, there is ${\pi }_0(hv)=\frac{1}{|G|}{\sum }_{g\in G}h(h^{-1}g)\pi (g^{-1}hv)=h{\pi }_0(v)=0$. Therefore, $\ker {\pi }_0$ is a $G$-submodule and we finish the proof. \end{proof}

Remark. Note that ${\pi }_0(hv)=h{\pi }_0(v)$ yields that ${\pi }_0:V\to W$ is a morphism of $G$-modules.

Example. Let $G=\mathbb{Z}$. Define a representation $\rho :G\to \mathrm{GL}({\mathbb{R}}^2),a\mapsto \left[\begin{array}{cc}1& a\\ & 1\end{array}\right]$. Then $G$ is a ${\mathbb{R}}^2$-module and $V=\{(x,0)\}\subset {\mathbb{R}}^2$ is a $G$-submodule. Then $W=\{(\alpha y,y):y\ne 0,\alpha \in \mathbb{R}\}$ is also a $G$-submodule and ${\mathbb{R}}^2=V\oplus W$.

Decomposition of Modules

Assume that $G$ is a finite group with any $G$-module $V=V_1^{a_1}\oplus \cdots \oplus V_k^{a_k}$, where $V_i$ is irreducible non-isomorphic modules and $a_i$ are multiplicities. Then by Schur’s lemma, this decomposition is unique. Yet we can not prove that there is a finite number of irreducible $G$-modules.

Example. Let $G$ be an abelian group $G=C_3=\{1,a,a^2\}$. If $\rho :G\to \mathrm{GL}(V)$ is a representation, then $\rho (h)\rho (g)=\rho (g)\rho (h)$ yields that $\rho (g):V\to V$ is a morphism of $G$-modules. If $V$ is irreducible and $k$ is a algebraically closed field, then by Schur’s lemma $\rho (g)v=\lambda v$ for some $\lambda \in k$. So any $v\in V$ is an eigenvector and $\rho (g)=\lambda I$. It follows that any subspace in $V$ is $G$-submodule and so $\dim V=1$ by $V$ irreducible. So we get the following proposition.

Proposition

Any irreducible representation of abelian group is one-dimensional.

• Therefore, if $\rho :C_3\to {\mathbb{C}}^{\ast }$ is a representation, then $\rho (a)\in \{1,\omega ,{\omega }^2\}$ where $\rho$ is a primitive root of unity of degree $3$.
• It is easy to verify that ${\rho }_1(a)=\omega$ and ${\rho }_2(a)={\omega }^2$ are non-isomorphic. In fact, we have that $|G|={\sum }_i\dim |V_i{|}^2$. So ${\rho }_i(a)={\omega }^i$ for $i=0,1,2$ are all irreducible $G$-module up to isomorphic.

Representations of $S_3$

In general, in $S_n$ permutation representation $\sigma$ in $k^n$, we have that $\sigma (e_i)=e_{\sigma (i)}$ and $\det (\rho (g))=\pm 1$. Note that $\mathrm{sgn}:S_n\to \{1,-1\}\subset C^{\ast },\sigma \mapsto \mathrm{sgn}\sigma$ is a one-dimensional representation.

For $S_3$ and its permutation representation $\rho :S_3\to \mathrm{GL}({\mathbb{C}}^3)$, we have two $S_3$-submodule $W=\{(\alpha ,\alpha ,\alpha ):\alpha \in \mathbb{C}\}\subseteq {\mathbb{C}}^3$ and $U=\{(\alpha ,\beta ,\gamma ):\alpha +\beta +\gamma =0\}$. Also, another $S_3$-module corresponding the representation $\mathrm{sgn}$ is defined above. Since $6=1^2+1^2+2^2$, all irreducible module of $S_3$ have been listed.

Let $U$ be an irreducible $S_3$-module. Let $\tau =(12)$ and $\sigma =(123)$. Then $S_3=⟨\tau ,\sigma ⟩$ and $C_3=\{1,\sigma ,{\sigma }^2\}⩽S_3$. Then $U$ is also a $C_3$-module. Since $C_3$ is abelian, $U$ can be written as $U=U_{\omega }^{a_1}+U_{{\omega }^2}^{a_2}+U_e^{a_3}$.

If $u$ is a eigenvector of $\sigma$ with eigenvalue ${\omega }^{\alpha }$, $\alpha \in \{0,1,2\}$, then $\sigma (\tau (u))=\tau ({\sigma }^2(u))=\tau ({\omega }^{2\alpha }u)={\omega }^{2\alpha }\tau (u)$, that is, $\tau (u)$ is an eigenvector of $\tau$ with eigenvalue ${\omega }^{2\alpha }$. If ${\omega }^{\alpha }\ne 1$, then ${\omega }^{2\alpha }\ne 1$. So $u,\tau (u)$ are linearly independent, and $⟨u,\tau (u)⟩$ is a $2$-dimensional $S_3$-module.

For example, we call $2$-dimensional $S_3$-module generated by $x=(\omega ,1,{\omega }^2)$ and $y=(1,\omega ,{\omega }^2)$ standard representation of $S_3$.

• $\sigma (x)=\omega x$
• $\sigma y={\omega }^{2}y$
• $\tau (x)=y$
• $\tau (y)=x$

Moreover, $x$ and $y$ is basis of $U=\{(\alpha ,\beta ,\gamma ):\alpha +\beta +\gamma =0\}$.

Definition

The group $S_n$ acting on ${\mathbb{C}}^n$ by permuting basis vectors defines a representation. $U:=\{a_1{e}_1+\cdots +a_n{e}_n:a_1+\cdots +a_n=0\}$ is a $S_n$-module, which is called the standard representation.

If $v_i$ and $\tau (v_i)$ are linear independent, then $v_i+\tau (v_i)$ are $v_i-\tau (v_i)$ are linear independent. Furthermore, $v_i+\tau (v_i)$ is a trivial representation while $v_i-\tau (v_i)$ is a sign representation.

If $v_i$ and $\tau (v_i)$ are linear dependent, then $⟨v_i,\tau (v_i)⟩$ is one-dimensional. Since ${\tau }^2(v_i)=v_i$, then $\tau (v_i)=\pm v_i$. If $\tau (v_i)=-v_i$, then $⟨v_i⟩$ is sign representation; otherwise $⟨v_i⟩$ is trivial.

Any finite-dimensional representation of $S_3$ is direct sum of trivial, sign and standard. Recall here.