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Let G be a finite group and let F be an algebraically closed field. If the characteristic of F is 0, then the number of irreducible F-representations of G is given by the number of conjugacy classes of elements of G. A paper I'm reading says that if the characteristic of F is p>0, then the number of F-irreps of G is the same as the number of conjugacy classes of elements whose order is not divisible by p.

If G is abelian, it seems to me that this should say that the p-sylow subgroup of G acts trivially on every characteristic p irrep. This is because I can split G into G'x P (non-p and p-sylow subgroups), and then any irrep of G' extends to one of G by letting P act trivially. Since the formula mentioned above would say that they have the same number of irreps these must be all of them.

My question is: Is this true? If not, then where is my reasoning going off track?

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Could you give the reference to this paper that you are reading, containing the claim that the number of F-irreps is the number of conjugacy classes of elements of order prime to p? – Leonid Positselski Jan 24 '10 at 22:06
'Iwasawa theory and projective modules' by Ralph Greenberg. – Joel Dodge Jan 24 '10 at 22:13
up vote 16 down vote accepted

Yes, you are correct. The point is that a $p$-group acting in char. $p$ always has a fixed point (and so acts trivially on an irrep.). So every irrep. of $G$ in char. $p$ factors through $G'$, as you anticipated.

The proof of the claim about $p$-groups is not hard. One approach (in general, even when $P$ is not nec. abelian) is to prove it by induction on the order of $P$, and so (using the fact that the descending central series is non-trivial) reduce to the case when $P$ is cyclic of order $p$. In this case, one can for example look at the group ring $k[P] = k[t]/(t^p - 1) = k[t-1]/(t-1)^p,$ and observe that it has a unique maximal ideal, and hence that $P$ has a unique irrep., namely the trivial one.

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Great thanks! If F is a finite field, then the claim about p-groups also follows from standard facts about groups acting on sets: If G is a p-group acting on a set X, then |X| = |X^G| (mod p). – Joel Dodge Jan 24 '10 at 19:56

The fact that the number of isomorphism types of irreducible $FG$ modules is the number of conjugacy classes of elements of $G$ of order prime to $p$ (or number of $p$-regular conjugacy classes) when $F$ is an algebraically closed field of characteristic $p$ is a theorem of Richard Brauer. Proofs can be found in almost any text which treats modular representations of finite groups (eg Curtis and Reiner,Representation Theory of Finite Groups and Associative Algebras, Wiley, 1962). It is a reasonably elementary exercise in finite group theory that, letting $O_{p}(G)$ denote the (unique) largest normal $p$-subgroup of $G$, there is a one-to-one correspondence between $p$-regular conjugacy classes of $G$ and $p$-regular conjugacy classes of $G/O_{p}(G)$. This is one way to see that $O_{p}(G)$ must act trivially on all irreducible $FG$-modules, a special case of which you had observed. A more representation theoretic way to see this is to note that if $V$ is an irreducible $FG$-module, then the subspace $V^{O_{p}(G)}$ of $O_{p}(G)$-fixed points on $V$ is a $G$-invariant supspace; as observed in comments above, this space is non-zero, so by irreducibility, it must be all of $V$. (This can be seen as a particular application of Clifford's Theorem).

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Brauer's proof that the number of similarity classes of irreducible representations of $G$ over an algebraically closed field of characteristic $p$ is equal to the number of $p$-regular conjugacy classes of $G$ is ring-theoretic in flavor, and rather tricky. There is also an easy character theoretic proof based on the following ideas. First, the set IBr$(G)$ of irreducible Brauer characters is in bijective correspondence with the irreducible representations, and this set of functions lives in the space $V$ of all complex-valued class functions defined on the set of p-regular elements. Since $\dim(V)$ equals the number of $p$-regular classes, it suffices to show that IBr($G$) is a basis for $V$. The linear independence of IBr$(G)$ is a standard result. To see that IBr$(G)$ spans, use the facts that Irr$(G)$ spans the space of all class functions and that on each $p$-regular class, the value of an ordinary character is a linear combination (and in fact, a sum) of values of Brauer characters.

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