Maschke's theorem

In mathematics, Maschke's theorem, named after Heinrich Maschke, is a theorem in group representation theory that concerns the decomposition of representations of a finite group into irreducible pieces. Maschke's theorem allows one to make general conclusions about representations of a finite group G without actually computing them. It reduces the task of classifying all representations to a more manageable task of classifying irreducible representations, since when the theorem applies, any representation is a direct sum of irreducible pieces (constituents). Moreover, it follows from the Jordan–Hölder theorem that, while the decomposition into a direct sum of irreducible subrepresentations may not be unique, the irreducible pieces have well-defined multiplicities. In particular, a representation of a finite group over a field of characteristic zero is determined up to isomorphism by its character.

Formulations
Maschke's theorem addresses the question: when is a general (finite-dimensional) representation built from irreducible subrepresentations using the direct sum operation? This question (and its answer) are formulated differently for different perspectives on group representation theory.

Group-theoretic
Maschke's theorem is commonly formulated as a corollary to the following result:

$$

Then the corollary is

$$

The vector space of complex-valued class functions of a group $$G$$ has a natural $$G$$-invariant inner product structure, described in the article Schur orthogonality relations. Maschke's theorem was originally proved for the case of representations over $$\Complex$$ by constructing $$U$$ as the orthogonal complement of $$W$$ under this inner product.

Module-theoretic
One of the approaches to representations of finite groups is through module theory. Representations of a group $$G$$ are replaced by modules over its group algebra $$K[G]$$ (to be precise, there is an isomorphism of categories between $$K[G]\text{-Mod}$$ and $$\operatorname{Rep}_{G}$$, the category of representations of $$G$$). Irreducible representations correspond to simple modules. In the module-theoretic language, Maschke's theorem asks: is an arbitrary module semisimple? In this context, the theorem can be reformulated as follows:

$$

The importance of this result stems from the well developed theory of semisimple rings, in particular, their classification as given by the Wedderburn–Artin theorem. When $$K$$ is the field of complex numbers, this shows that the algebra $$K[G]$$ is a product of several copies of complex matrix algebras, one for each irreducible representation. If the field $$K$$ has characteristic zero, but is not algebraically closed, for example if $$K$$ is the field of real or rational numbers, then a somewhat more complicated statement holds: the group algebra $$K[G]$$ is a product of matrix algebras over division rings over $$K$$. The summands correspond to irreducible representations of $$G$$ over $$K$$.

Category-theoretic
Reformulated in the language of semi-simple categories, Maschke's theorem states

$$

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

Define $p(x) = \frac{1}{\#G} \sum_{g \in G} g \cdot p_0 \cdot g^{-1} (x)$, where $$g \cdot p_0 \cdot g^{-1}$$ is an abbreviation of $$\rho_W{g} \cdot p_0 \cdot \rho_V{g^{-1}}$$, with $$\rho_W{g}, \rho_V{g^{-1}}$$ being the representation of G on W and V. Then, $$\ker p$$ is preserved by G under representation $$\rho_V$$: for any $$w' \in \ker p, h \in G$$, $$\begin{align} p(hw') &= h \cdot h^{-1} \frac{1}{\#G} \sum_{g \in G} g \cdot p_0 \cdot g^{-1} (hw') \\ &= h \cdot \frac{1}{\#G} \sum_{g \in G} (h^{-1} \cdot g) \cdot p_0 \cdot (g^{-1} h) w' \\ &= h \cdot \frac{1}{\#G} \sum_{g \in G} g \cdot p_0 \cdot g^{-1} w' \\ &= h \cdot p(w') \\ &= 0 \end{align}$$

so $$w' \in \ker p$$ implies that $$hw' \in \ker p $$. So the restriction of $$\rho_V$$ on $$\ker p$$ is also a representation.

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

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 $$ \begin{cases} \varphi:K[G]\to V \\ \varphi:x \mapsto \frac{1}{\#G}\sum_{s \in G} s\cdot \pi(s^{-1} \cdot x) \end{cases}$$

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

$$\begin{align} \varphi(t\cdot x) &= \frac{1}{\#G}\sum_{s \in G} s\cdot \pi(s^{-1}\cdot t\cdot x)\\ &= \frac{1}{\#G}\sum_{u \in G} t\cdot u\cdot \pi(u^{-1}\cdot x)\\ &= t\cdot\varphi(x), \end{align}$$

so φ is in fact K[G]-linear. By the splitting lemma, $$K[G]=V \oplus \ker \varphi$$. 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 $\epsilon(x) = \sum\lambda_g$. Let $$I=\ker\epsilon$$. Then I is a K[G]-submodule. We will prove that for every nontrivial submodule V of K[G], $$I \cap V \neq 0$$. Let V be given, and let $v=\sum\mu_gg$ be any nonzero element of V. If $$\epsilon(v)=0$$, the claim is immediate. Otherwise, let $s = \sum 1 g$. Then $$\epsilon(s) = \#G \cdot 1 = 0$$ so $$s \in I$$ and $$sv = \left(\sum1g\right)\!\left(\sum\mu_gg\right) = \sum\epsilon(v)g = \epsilon(v)s$$

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