Let $K$ be a field and $G$ be a finite group, and denote by $K[G]$ its group ring, then Maschke's theorem is:

Suppose $\mbox{char}(K)$ does not divide $|G|$. Let $V$ be a $K[G]$-module. If $W \subseteq V$ is a $K[G]$-submodule, then there is a submodule $W' \subseteq V$ such that: $$ W + W' = V, \quad W \cap W' = 0. $$ Proof: In the proof by linear algebra, we can choose a linear subspace (where $V$ is considered as a $K$-vector space) $X$ of $V$ such that $V = W + X, W\cap X = 0$. Then we take the projection $\pi : V \to W$, which in general is not a $K[G]$-module homomorphism, so we modify it to $$ q(v) := \frac{1}{|G|} \sum_{g\in G} g(\pi(g^{-1}v)) $$ and this gives an $K[G]$-module homomorphism. Then $W' = \mbox{ker}(q)$ (I omit the verficational part of the proof). $\square$

This proof I found in many sources (for example here, here and here, or also in S. Lang's book Algebra) . But what I do not understand is the multiplication by $1/|G| \in \mathbb Q$. In general, the sum $$ \sum_{g\in G} g(\pi(g^{-1}v)) \in W \subseteq V $$ as it is a $K[G]$-submodule, but how could we multiply an element from $W$ (or $V$) by a rational number $1/|G|$? That does not make much sense to me...


The reason for rescaling is so that $q(q(v))=q(v)$ (i.e. $q$ is a projection operator). The number $1/|G|$ is not a rational number. Instead $1/|G|=|G|^{-1}\in K$, which exists since by assumption $0\neq |G|=|G|\cdot 1_K\in K$.

  • $\begingroup$ Ah okay, thats the reason for the assumption of the characteristic. By the way, if $\mbox{char}(K) = 0$ we can always embed $\mathbb N$ into $K$ by $n \equiv 1_K + \ldots + 1_K$ ($n$ times) [injective as otherwise $\mbox{char} \ne 0$], I guess with $a/b \equiv a\cdot b^{-1}$ there would be a similar embedding of $\mathbb Q$ in $K$. Okay, a general theorem is that the prime field is isomorphic to $\mathbb Q$, so this fact seems to be well-known. But what about $\mathbb R$, could we build Cauchy sequence and get $\mathbb R\hookrightarrow K$ (an embedding of $\mathbb R$) if $K \ncong \mathbb Q$? $\endgroup$ – StefanH Oct 13 '15 at 16:46
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    $\begingroup$ You are overcomplicating this. There is a ring homomorphism $\mathbb{Z}\to K$ given by $1\mapsto 1_K$. If this map is injective, then using the universal property of quotient fields $\mathbb{Q}\hookrightarrow K$. Otherwise, the kernel is $n\mathbb{Z}$ for some $n$ and $\mathbb{Z}/n\mathbb{Z}\hookrightarrow K$. Since subrings of fields are domains, we conclude $n=p$ is prime, so $\mathbb{Z}_p\subset K$. You cannot guarantee $\mathbb{R}$ is a subfield of $K$, even when $K= \overline{K}$ is of characteristic 0. $\endgroup$ – David Hill Oct 13 '15 at 16:51
  • $\begingroup$ Thanks, that clarified a lot! Do you have an example of a field of characteristic $0$ which properly contains its prime field, but does not contain a copy of $\mathbb R$? And by the way, what is $\overline K$? $\endgroup$ – StefanH Oct 13 '15 at 16:58
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    $\begingroup$ $\overline{K}$ is the algebraic closure of $K$ (so all polynomials split). For your other question, take $\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}\mid a,b\in\mathbb{Q}\}$. If you want algebraically closed, the algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$ is countable (hence, doesn't contain $\mathbb{R}$. $\endgroup$ – David Hill Oct 13 '15 at 17:02

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