TITLE: Why should we expect Maschke's theorem to be true?
QUESTION [11 upvotes]: Maschke's theorem tells us that any representation of a finite group $G$ can be decomposed into a direct sum of irreducible representations. The proof does make intuitive sense to me (Intuition behind Maschke's theorem), but my question is really about why we should expect it to be true in the first place.
Sure, we can do some explicit examples (perhaps a most obvious starting point is the standard permutation representation of $S_n$, which has an obvious invariant subspace $\{(x,x,\ldots, x) \in \mathbb C^n: x\in \mathbb C\}$, and almost as obvious complement subspace $\{(x_1,\ldots, x_n) \in \mathbb C^n: \sum x_i = 0\}$), and perhaps one has already seen the result for abelian groups over $\mathbb C$ in the guise of linear algebra: Matrices commute if and only if they share a common basis of eigenvectors?.
However I wonder if there is any other point of view that makes Maschke's theorem feel "inevitable"...since right now, it just seems absurdly powerful/magical.

REPLY [4 votes]: As sort of an anti-answer, why should we expect it to be true? Isn't it a bit surprising when you first see it?
Just look at representations of the most basic infinite group, the infinite cyclic group $\mathbb{Z}$. The representations correspond to matrices, and their decomposition corresponds to the Jordan normal form. For example, the representation $$f: (\mathbb{Z},+) \rightarrow GL_2(\mathbb{C}),\ f(x) = \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix}$$ is not completely reducible.
So what about finite groups? Look at the simplest example, a cyclic group $G = \langle g \rangle$, say $|g| = n$. For a representation $f: G \rightarrow \operatorname{GL}_d(\mathbb{C})$, you have $f(g)^n = I_d$. So the minimal polynomial of $f(g)$ divides $x^n - 1$, which is a polynomial with no repeated roots. Hence $f(g)$ is diagonalizable, meaning after a change of basis $f(g)$ is a diagonal matrix $$f(g) = \begin{pmatrix} \lambda_1 & & & \\ & \lambda_2 & & \\ & & \ddots & \\ & & & \lambda_d \end{pmatrix}.$$
Hence representations of finite cyclic groups are completely reducible.
But why should we expect representations of all finite groups to be completely reducible? I do not think it is obvious at all a priori.
And also it's not true over fields of positive characteristic, here is one example: take a finite cyclic group $G = \langle g \rangle$ of prime order $p$. Then $$f: G \rightarrow \operatorname{GL}_2(\mathbb{Z} / p\mathbb{Z}), f(g^k) = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$$ is a representation which is not completely reducible.
So you somehow need to use the fact that the group is finite, and the fact that the field is of characteristic zero (or more generally, of characteristic coprime to the order of the group). As shown in the other answers, a certain averaging argument does the trick.