Suppose $G$ is a finite group, and $l$ is a prime, with $l$ coprime to the order $G$. (Thus we have complete reducibility for $G$ representations.) Is there a straightforward condition on $l$ which ensures that every irreducible representation of $G$ is liftable to a characteristic zero representation? (For instance, does the fact that we assume $l$ coprime to $G$ suffice?)

$\begingroup$ Do we need a new tag, characteristicl? :) $\endgroup$ – Pete L. Clark Jan 24 '10 at 22:16
Yes, $\operatorname{gcd}(l,G) = 1$ is sufficient. This is an easy consequence of Brauer's modular representation theory. See Serre's Linear Representations of Finite Groups, especially Chapter 18.

$\begingroup$ By the way, this reference is somehow overkill. It would be nice to see a more direct argument in this much easier special case, e.g. starting with a version of Maschke's Theorem over $\mathbb{Z}_{\ell}[G]$. My instruments predict a response coming in from a northwesternly direction... $\endgroup$ – Pete L. Clark Jan 24 '10 at 22:41

$\begingroup$ Use the splitting of $A \rightarrow G \rightarrow B$ whenever the orders of $A$ and $B$ are relatively prime. $\endgroup$ – moonface Jan 25 '10 at 1:19

3$\begingroup$ One possible argument is via deformation theory, which will be very natural for some (but maybe not all). Namely, given the irrep. $\rho$ in char. $\ell$, the obstructions to deformation are computed by ${\mathrm Ext}^2_{{\mathbb F}\_{\ell}}(\rho,\rho),$ and the mod $\ell$ tangent space to the deformation space is computed by ${\mathrm Ext}^1_{{\mathbb F}\_{\ell}}(\rho,\rho).$ Since both vanish when $\ell$ is prime to $G$, we see that each irrep. $\rho$ mod $\ell$ lifts uniquely to an irrep. in char. 0. $\endgroup$ – Emerton Jan 25 '10 at 2:29

$\begingroup$ Typo: in the above, the subscript ${\mathbb F}_{\ell}$ on the Ext spaces should be ${\mathbb F}_{\ell}[G]$. $\endgroup$ – Emerton Jan 25 '10 at 2:30
Another sufficient condition is that if $G$ is solvable, then for every prime $l$, every absolutely irreducible characteristic $l$ representation can be lifted to the complex numbers. In fact, solvability is not really necessary; $l$solvability suffices. This is the FongSwan theorem.
Added later: Since groups with order not divisible by $l$ are trivially $l$solvable, this sufficient condition includes, and is more general than the the condition stated by Pete L. Clark.