# Maschke's theorem

In mathematics, Maschke's theorem,[1][2] named after Heinrich Maschke,[3] 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:

Theorem. If V is a complex representation of a finite group G with a subrepresentation W, then there is another subrepresentation U of V such that V=WU.[4][5]

Then the corollary is

Corollary (Maschke's theorem). Every representation of a finite group G over a field F with characteristic not dividing the order of G is a direct sum of irreducible representations.[6][7]

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 ${\displaystyle \mathbb {C} }$  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]-Mod and RepG, 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:

Maschke's Theorem. Let G be a finite group and K a field whose characteristic does not divide the order of G. Then K[G], the group algebra of G, is semisimple.[8][9]

The importance of this result stems from the well developed theory of semisimple rings, in particular, the Artin–Wedderburn theorem (sometimes referred to as Wedderburn's Structure 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.[10] If the field K has characteristic zero, but is not algebraically closed, for example, K is a 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.[11]

### Category-theoretic

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

Maschke's theorem. If G is a group and F is a field with characteristic not dividing the order of G, then the category of representations of G over F is semi-simple.

## 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 ${\displaystyle p(x)={\frac {1}{\#G}}\sum _{g\in G}g\cdot p_{0}\cdot g^{-1}(x)}$ , 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 ${\displaystyle w'\in ker\ p,h\in G}$ ,

{\displaystyle {\begin{aligned}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{aligned}}}

so ${\displaystyle w'\in ker\ p}$  implies that ${\displaystyle hw'\in ker\ p}$ . 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

${\displaystyle {\begin{cases}\phi :K[G]\to V\\\phi (x)={\frac {1}{\#G}}\sum _{s\in G}s\cdot \pi (s^{-1}\cdot x)\end{cases}}}$

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

{\displaystyle {\begin{aligned}\phi (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 \phi (x),\end{aligned}}}

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.[12]

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

${\displaystyle sv=\left(\sum 1g\right)\left(\sum \mu _{g}g\right)=\sum \epsilon (v)g=\epsilon (v)s}$

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 ${\displaystyle \rho :\mathbb {Z} \to GL_{2}(\mathbb {C} )}$  defined by ${\displaystyle \rho (n)={\begin{bmatrix}1&1\\0&1\end{bmatrix}}^{n}={\begin{bmatrix}1&n\\0&1\end{bmatrix}}}$ . Let ${\displaystyle W=\mathbb {C} \cdot {\begin{bmatrix}1\\0\end{bmatrix}}}$ , a 1-dimensional subspace of ${\displaystyle GL_{2}(\mathbb {C} )}$  spanned by ${\displaystyle {\begin{bmatrix}1\\0\end{bmatrix}}}$ . 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 ${\displaystyle \mathbb {Z} =W\oplus U}$ : any such U needs to be 1-dimensional, but any 1-dimensional subspace preserved by ${\displaystyle \rho }$  has to be spanned by eigenvector for ${\displaystyle {\begin{bmatrix}1&1\\0&1\end{bmatrix}}}$ , and the only eigenvector for that is ${\displaystyle {\begin{bmatrix}1\\0\end{bmatrix}}}$ .
• Consider a prime p, and the group ${\displaystyle \mathbb {Z} /_{p}\mathbb {Z} }$ , field ${\displaystyle K=\mathbb {F} _{p}}$ , and the representation ${\displaystyle \rho :\mathbb {Z} /_{p}\mathbb {Z} \to GL_{2}(\mathbb {F} _{p})}$  defined by ${\displaystyle \rho (n)={\begin{bmatrix}1&n\\0&1\end{bmatrix}}}$ . Simple calculations show that there is only one eigenvector for ${\displaystyle {\begin{bmatrix}1&1\\0&1\end{bmatrix}}}$  here, so by the same argument, the 1-dim 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.

## Notes

1. ^ Maschke, Heinrich (1898-07-22). "Ueber den arithmetischen Charakter der Coefficienten der Substitutionen endlicher linearer Substitutionsgruppen" [On the arithmetical character of the coefficients of the substitutions of finite linear substitution groups]. Math. Ann. (in German). 50 (4): 492–498. doi:10.1007/BF01444297. JFM 29.0114.03. MR 1511011.
2. ^ Maschke, Heinrich (1899-07-27). "Beweis des Satzes, dass diejenigen endlichen linearen Substitutionsgruppen, in welchen einige durchgehends verschwindende Coefficienten auftreten, intransitiv sind" [Proof of the theorem that those finite linear substitution groups, in which some everywhere vanishing coefficients appear, are intransitive]. Math. Ann. (in German). 52 (2–3): 363–368. doi:10.1007/BF01476165. JFM 30.0131.01. MR 1511061.
3. ^
4. ^ Fulton & Harris, Proposition 1.5.
5. ^ Serre, Theorem 1.
6. ^ Fulton & Harris, Corollary 1.6.
7. ^ Serre, Theorem 2.
8. ^ It follows that every module over K[G] is a semisimple module.
9. ^ The converse statement also holds: if the characteristic of the field divides the order of the group (the modular case), then the group algebra is not semisimple.
10. ^ The number of the summands can be computed, and turns out to be equal to the number of the conjugacy classes of the group.
11. ^ One must be careful, since a representation may decompose differently over different fields: a representation may be irreducible over the real numbers but not over the complex numbers.
12. ^ Serre, Exercise 6.1.