Maschke's theorem

Proofs

Group-theoretic

Let U be a subspace of V complement of W. Let ${\displaystyle p_0:V\to W}$ be the projection function, i.e., ${\displaystyle p_0(w+u)=w}$ for any ${\displaystyle u\in U,w\in W}$.

Define $p(x)=\frac{1}{\mathrm{\#}G}\sum _{g\in G}g\cdot p_0\cdot g^{-1}(x)$, where ${\displaystyle g\cdot p_0\cdot g^{-1}}$ is an abbreviation of ${\displaystyle {\rho }_{W}g\cdot p_0\cdot {\rho }_V{g}^{-1}}$, with ${\displaystyle {\rho }_{W}g,{\rho }_V{g}^{-1}}$ being the representation of G on W and V. Then, ${\displaystyle \ker p}$ is preserved by G under representation ${\displaystyle {\rho }_V}$: for any ${\displaystyle w^{\prime }\in \ker p,h\in G}$, $${\displaystyle \begin{array}{cc}p(h{w}^{\prime })& =h\cdot h^{-1}\frac{1}{\mathrm{\#}G}\sum _{g\in G}g\cdot p_0\cdot g^{-1}(h{w}^{\prime })\\ & =h\cdot \frac{1}{\mathrm{\#}G}\sum _{g\in G}(h^{-1}\cdot g)\cdot p_0\cdot (g^{-1}h)w^{\prime }\\ & =h\cdot \frac{1}{\mathrm{\#}G}\sum _{g\in G}g\cdot p_0\cdot g^{-1}{w}^{\prime }\\ & =h\cdot p(w^{\prime })\\ & =0\end{array}}$$

so ${\displaystyle w^{\prime }\in \ker p}$ implies that ${\displaystyle h{w}^{\prime }\in \ker p}$. So the restriction of ${\displaystyle {\rho }_V}$ on ${\displaystyle \ker p}$ is also a representation.

By the definition of ${\displaystyle p}$, for any ${\displaystyle w\in W}$, ${\displaystyle p(w)=w}$, so ${\displaystyle W\cap \ker p=\{0\}}$, and for any ${\displaystyle v\in V}$, ${\displaystyle p(p(v))=p(v)}$. Thus, ${\displaystyle p(v-p(v))=0}$, and ${\displaystyle v-p(v)\in \ker p}$. Therefore, ${\displaystyle V=W\oplus \ker p}$.

Module-theoretic

Let V be a K[G]-submodule. We will prove that V is a direct summand. Let π be any K-linear projection of K[G] onto V. Consider the map $${\displaystyle \{\begin{array}{l}\phi :K[G]\to V\\ \phi :x\mapsto \frac{1}{\mathrm{\#}G}\sum _{s\in G}s\cdot \pi (s^{-1}\cdot x)\end{array}}$$

Then φ is again a projection: it is clearly K-linear, maps K[G] to V, and induces the identity on V (therefore, maps K[G] onto V). Moreover we have

$${\displaystyle \begin{array}{cc}\phi (t\cdot x)& =\frac{1}{\mathrm{\#}G}\sum _{s\in G}s\cdot \pi (s^{-1}\cdot t\cdot x)\\ & =\frac{1}{\mathrm{\#}G}\sum _{u\in G}t\cdot u\cdot \pi (u^{-1}\cdot x)\\ & =t\cdot \phi (x),\end{array}}$$

so φ is in fact K[G]-linear. By the splitting lemma, ${\displaystyle K[G]=V\oplus \ker \phi }$. This proves that every submodule is a direct summand, that is, K[G] is semisimple.

Converse statement

The above proof depends on the fact that #G is invertible in K. This might lead one to ask if the converse of Maschke's theorem also holds: if the characteristic of K divides the order of G, does it follow that K[G] is not semisimple? The answer is yes.

Proof. For $x=\sum {\lambda }_{g}g\in K[G]$ define $ϵ(x)=\sum {\lambda }_g$. Let ${\displaystyle I=\ker ϵ}$. Then I is a K[G]-submodule. We will prove that for every nontrivial submodule V of K[G], ${\displaystyle I\cap V\ne 0}$. Let V be given, and let $v=\sum {\mu }_{g}g$ be any nonzero element of V. If ${\displaystyle ϵ(v)=0}$, the claim is immediate. Otherwise, let $s=\sum 1g$. Then ${\displaystyle ϵ(s)=\mathrm{\#}G\cdot 1=0}$ so ${\displaystyle s\in I}$ and $${\displaystyle sv=\left(\sum 1g\right)\left(\sum {\mu }_{g}g\right)=\sum ϵ(v)g=ϵ(v)s}$$

so that ${\displaystyle sv}$ is a nonzero element of both I and V. This proves V is not a direct complement of I for all V, so K[G] is not semisimple.

Non-examples

The theorem can not apply to the case where G is infinite, or when the field K has characteristics dividing #G. For example,

• Consider the infinite group ${\displaystyle \mathbb{Z}}$ and the representation ${\displaystyle \rho :\mathbb{Z}\to {\mathrm{G}\mathrm{L}}_2(\mathbb{C})}$ defined by ${\displaystyle \rho (n)={\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]}^n=\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]}$. Let ${\displaystyle W=\mathbb{C}\cdot \left[\begin{array}{c}1\\ 0\end{array}\right]}$, a 1-dimensional subspace of ${\displaystyle {\mathbb{C}}^2}$ spanned by ${\displaystyle \left[\begin{array}{c}1\\ 0\end{array}\right]}$. Then the restriction of ${\displaystyle \rho }$ on W is a trivial subrepresentation of ${\displaystyle \mathbb{Z}}$. However, there's no U such that both W, U are subrepresentations of ${\displaystyle \mathbb{Z}}$ and ${\displaystyle {\mathbb{C}}^2=W\oplus U}$: any such U needs to be 1-dimensional, but any 1-dimensional subspace preserved by ${\displaystyle \rho }$ has to be spanned by an eigenvector for ${\displaystyle \left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]}$, and the only eigenvector for that is ${\displaystyle \left[\begin{array}{c}1\\ 0\end{array}\right]}$.
• Consider a prime p, and the group ${\displaystyle \mathbb{Z}/p\mathbb{Z}}$, field ${\displaystyle K={\mathbb{F}}_p}$, and the representation ${\displaystyle \rho :\mathbb{Z}/p\mathbb{Z}\to {\mathrm{G}\mathrm{L}}_2({\mathbb{F}}_p)}$ defined by ${\displaystyle \rho (n)=\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]}$. Simple calculations show that there is only one eigenvector for ${\displaystyle \left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]}$ here, so by the same argument, the 1-dimensional subrepresentation of ${\displaystyle \mathbb{Z}/p\mathbb{Z}}$ is unique, and ${\displaystyle \mathbb{Z}/p\mathbb{Z}}$ cannot be decomposed into the direct sum of two 1-dimensional subrepresentations.