Why does Maschke's theorem require the characteristic of the field to be coprime to the order of the group? I was reading a proof of Maschke's theorem (specifically pages 5-6 of this paper) and it seemed relatively straightforward... the only problem is the extra condition of coprimeness, which doesn't seem necessary to me. Where does the proof fail if the coprime condition is not satisfied? Also, what happens for fields of characteristic 0, such as the complex numbers? Isn't that not coprime to the order of any group except the trivial group?
 A: The normal reason in representation theory, as here, is that some of the arguments involve an averaging process, which involves dividing by the order of the group. This is not possible if the order is not invertible in the ground field.
Of course this is automatically possible in characteristic $0$, which is not taken to be excluded by the coprime condition in this context (by convention, if you like).
[Aside to be ignored if confusing: JH Conway uses $-1$ as the "prime" associated with characteristic $0$ in his book on quadratic forms]
A: "Coprime" is maybe a confusing way to state it for consistency with the characteristic zero case; the condition we need is that $|G|$ is invertible over $k$, and as the other answers say it's so that we can divide by it (which in characteristic $0$ we always can), which the author does on page 6.
This is the sort of thing that can happen if $|G|$ isn't invertible. Take $k = \mathbb{F}_p, G = C_p$ and consider the $2$-dimensional representation
$$C_p \ni k \mapsto \left[ \begin{array}{cc} 1 & k \\ 0 & 1 \end{array} \right]$$
which I invite you to check has a $1$-dimensional invariant subspace without an invariant complement. (It is a minor crime to prove Maschke's theorem and not mention this counterexample.)
A: As you can see in the proof of Maschke's theorem, you need the inverse of the group order, i.e., $\frac{1}{|G|}$. And this number only exists if the characteristic of the underlying field is not a divisor of $|G|$.
