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Let $K$ be a field and $G$ a finite group such that $\text{ch}(K)\nmid |G|$. We have defined the group ring $$K[G]:=\left\{\sum_{g\in G}k_g\chi_g:k_g\in K,k_g\neq0\text{ for only finitely many }g\in G\right\}$$ via the addition of functions and the convolution product, i.e. the $K$-bilinear extension of $\chi_g\ast\chi_h=\chi_{gh}$. Here $\chi_g$ means the characteristic function for $g\in G$.

With the assumption $\text{ch}(K)\nmid |G|$ I want to prove the

Claim. $K[G]$ is semi-simple (as a ring), i.e. the left $K[G]$-module $_{K[G]}K[G]$ is a direct sum of simple submodules.

Remark. The conditions 1) $_{K[G]}K[G]$ is semi-simple, 2) $_{K[G]}K[G]$ is a sum of simple submodules and 3) for every submodule $N\subset _{K[G]}K[G]$ there is a submodule $L$ such that $_{K[G]}K[G]=N\oplus L$ are all equivalent.

We have to work with the

Hint. Let $V$ be a $K[G]$-module and $U$ a submodule of $V$. Since $V$ is a $K$-vectorspace, we can find a subvectorspace $W$ of $V$ such that $V=U\oplus W.$ Let $P$ be the projection onto $U$ and $$\widetilde{P}:=\frac{1}{|G|}\sum_{g\in G}\rho(g)P\rho(g^{-1}),$$ where $\rho(g)$ is the action of $g$ on $V$. Now show that $\widetilde{P}$ is a $K[G]$-module homomorphism and $\widetilde{P}^2=\widetilde{P}$.

Now assume that I can show the two points in the hint. How does the claim follow?

2 Answers 2

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If you let $U'=\ker\widetilde{P}$, those two points will show that $\widetilde{P}$ is a projection along $U'$, so that $V=U\oplus U'$ as vector spaces. One checks again that $U'$ is stable under the $G$-action, hence is actually a $K[G]$-submodule, and thus $V=U\oplus U'$ is actually a decomposition into $K[G]$-submodules.

This shows any finite dimensional representation of $K[G]$ is completely reducible, which is one of the definitions of $K[G]$ being semisimple.

Perhaps you are aware, but this is Maschke's Theorem, details of the above can be found in Etingof's Represenation Theory, particularly Proposition 3.5.8 and Theorem 4.1.1.

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    Thank you! I just have three questions: 1) Why do we write $\rho(g)v$ and not simple $gv$? $V$ is a $K[G]$-module and $G\subset K[G]$, so $gv$ makes sense to me. 2) The result follows by my remark in the question, equivalence 3 with $V=K[G]$, right? 3) It is right that if we have $v\in V$ with $v=u+u'$ (recall $V=U\oplus U'$) then we have $\widetilde{P}(v)=u$? – 2017-01-04
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    @mathmarseille A $G$-module can also be thought of as a homomorphism $\rho\colon G\to GL(V)$. So $\rho(g)$ is an endomorphism of $V$. Many times we write $\rho(g)(v)=g\cdot v$ for ease of notation. Then the usual group action properties $(g_1g_2)\cdot v=g\cdot(h\cdot v)$ and $e\cdot v=v$ are just the homomorphism properties of $\rho(g_1g_2)=\rho(g_1)\rho(g_2)$ and $\rho(e)=id_V$. So it's fine to think of $gv$ instead of $\rho(g)(v)$. You're correct on the other two questions. – 2017-01-05
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    Thank you! But this is strange: If we really have $\widetilde{P}(u+u')=u$, then $\widetilde{P}=P$. But I guess this is not true. – 2017-01-05
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    @mathmarseille I don't think so, $\widetilde{P}$ is the projection onto $U$ along $U'$, whereas $P$ is the projection onto $U$ along $W$, so they will possibly be slightly different. – 2017-01-06
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If $M$ is an $R$-module and $p \colon M \to M$ is an idempotent endomorphism (i.e. $p^2 = p$) then $M = \operatorname{im} p \oplus \ker p$ (this is a straightforward calculation). (The mapping $p \mapsto (\operatorname{im} p, \ker p)$ does in fact give a bijection between the idempotent endomorphisms of $M$ and the pairs $(L,N)$ of submodules $L, N \subseteq M$ with $M = L \oplus N$.)

Given any submodule $U \subseteq V$ the hint shows how to construct an idempotent endomorphism $\tilde{P} \colon V \to V$ with $\operatorname{im} \tilde{P} = U$. Then $V = \ker \tilde{P} \oplus \operatorname{im} \tilde{P} = \ker \tilde{P} \oplus U$. Hence every submodule of $V$ has a direct complement.

If $V$ is finite-dimensional then by iterating this process we can decompose $V$ into a direct sum of simple modules. This does in particular apply to ${}_{K[G]} K[G]$ itself, since $G$ is finite.

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    Thank you! The only missing point is now the calculation of $\widetilde{P}^2=\widetilde{P}$. Can you say something about this? I don't get the right result. – 2017-01-04
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    Here it is probably easiest to check that $\tilde{P}$ fixes the elements $U$ pointwise (i.e. $\tilde{P}(u) = u$ for every $u \in U$) and that $\operatorname{im} \tilde{P} = U$. – 2017-01-04
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    Where have we used the structure of $K[G]$? For example the special multiplication? I guess we have only used the multiplication in $G$. – 2017-01-04