5
$\begingroup$

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?

$\endgroup$
1

3 Answers 3

7
$\begingroup$

"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.)

$\endgroup$
6
  • 1
    $\begingroup$ I don't think an integer $n>1$ is normally considered coprime with $0$ (since $n$ divides both of them). Therefore the author must have bent the notion of coprime when it comes to characteristics in the first place in order to make this a sensible statement, and using it "for the sake of consistency" seems a rather weak bid. To say that $|G|$ is "invertible over $k$" (or in $k$) or "not divisible by the characteristic" seem easier and clearer ways to express what is intended. $\endgroup$ Sep 15, 2020 at 15:48
  • 1
    $\begingroup$ @Marc: we are in agreement! $\endgroup$ Sep 15, 2020 at 16:20
  • $\begingroup$ Oh, thank you! I am so used to having scalars be complex numbers that I didn't even think twice about that step. What happens if the field used does not include integers? Then |G| is not an element of the field at all... $\endgroup$ Sep 15, 2020 at 21:05
  • 2
    $\begingroup$ @Isaiah: every ring $R$ has a canonical map $\mathbb{Z} \to R$ given by taking multiples of the identity, and it's an extremely common abuse of notation for people to write an integer $n$ as if it were an element of a ring $R$; they mean the element $n \cdot 1_R = \underbrace{1_R + \dots + 1_R}_{n \text{ times}}$ where $1_R$ is the identity. In positive characteristic this map has nontrivial kernel. $\endgroup$ Sep 15, 2020 at 21:09
  • 1
    $\begingroup$ This is also a nice example because it shows how Maschke can fail for infinite groups as well. $\endgroup$ Sep 19, 2020 at 21:16
6
$\begingroup$

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]

$\endgroup$
2
  • $\begingroup$ Interesting. Why does Conway do that? $\endgroup$
    – Elle Najt
    Sep 15, 2020 at 7:24
  • 1
    $\begingroup$ @LorenzoNajt I am sure there are various reasons, but one is using a multiplicative unit rather than the zero element goes through to various formulae, and it enables him to deploy the sign in a multiplicative context. As noted in the question here, using zero is not wholly smooth. $\endgroup$ Sep 15, 2020 at 11:16
4
$\begingroup$

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 $|G|$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .