That is: is it true that if projective k[G]-modules have same composition factors then they are isomorphic?

This is easy to see for char(k)=0, or if G is a composition of a p-group and a p′-group. Serre in "Linear Representations of Finite Groups" (a remark in 16.2 after Corr.2) states this as a well-known fact: "Indeed we know that the equality [P]=[P′] ... is equivalent to P=P′". But unfortunately no references.

  • $\begingroup$ I edited tags. Note too that your equal sign in the quote from Serre should be the isomorphism symbol. $\endgroup$ – Jim Humphreys Dec 10 '12 at 13:20

Some clarifications to the question are needed. First, you are referring to Section 16.1 of Serre's book (not 16.2), where he is formulating the main results in Brauer theory. These were originally derived (as in the 1962 Curtis-Reiner text) more concretely in terms of Brauer characters, but then recast in the language of Grothendieck groups. The result you are asking about is formally stated by Serre as Corollary 2 of Theorem 35 in that same section.

Serre's parenthetic remark after Corollary 2 of Theorem 34 in 16.1 refers back to the earlier and more elementary step in his Corollary 2 of Proposition 42 in 14.4.

There are two essential steps here, based always on the fact that you have a triple $(K,A,k)$ involving a residue field $k$ of characteristic $p>0$ coming from a suitable ring $A$ whose field of fractions $K$ has characteristic 0. (The deeper results require some hypotheses on completeness, splitting fields, etc.) First you compare the Grothendieck groups of projective modules for $KG$ and $AG$ (as in Serre's Section 14). Then you build the $cde$-triangle as in Section 16 and obtain there the injectivity of the map $c$ as Geoff indicates. (The later book Methods of Representation theory I by Curtis-Reiner has a more detailed version of all this theory.)

Let me mention that working with a finite group is essential here, since a partial parallel exists for certain finite group schemes (in the guise of restricted Lie algebras) for which the matrix of Cartan invariants has determinant 0 and the behavior of projectives is more complicated.

  • $\begingroup$ Jim - yes, 16.1, my bad. I still don't get the arugment. Corr. 2 of Prop. 42 in 14.4 says: P=Q if [P]=[Q] in $$ P_A[G] $$, not in $$ R_A[G] $$. That is: if P and Q have same indecoposable projective factors, then they are isomorphic (which is already stated in the preceding Corr 1.). From this I can't deduct that if P and Q have same irreducible factors, then they are isomophic. As for using the cde-triangle: Serre uses 16.1 as a building step to prove cde properties in Ch. 17, so I am not sure that using it here wouldn't create a cycled reference. Thanks for the ref to Curtis-Reiner. $\endgroup$ – George Dec 11 '12 at 4:00
  • $\begingroup$ I think what I wrote is accurate: your original question gets answered only by Corollary 2 of Theorem 35 in 16.1. (Section 14 is more elementary and preliminary.) Section 16.1 is very concise, so you have to follow up with details of Serre's proofs later on. His formalism is efficient but it takes real work to identify the main points in the proofs and their logical dependence. You might find the concrete Brauer style in the older Curtis-Reiner book to be more attractive. But the theorems are not easy in any case. Good luck. $\endgroup$ – Jim Humphreys Dec 11 '12 at 18:28
  • $\begingroup$ Thanks, Jim, Geoff. Unfortunately, I am not allowed to accept both answers (or I don't know how), so accepting from Jim. $\endgroup$ – George Dec 11 '12 at 20:11

This is true for finite groups, and it is a consequence of the non-singularity of the Cartan matrix (whose determinant is a power of $p$) in the algebraically closed case. The Cartan invariant $c_{ij}$ gives the multiplicity of the $j$-th simple module as a composition factor of the $i$-th projective indecomposable. If there were two non-isomorphic projectives with the same composition factors, the Cartan matrix would certainly be non-singular. I believe the result may have been stated by R. Swan.

  • $\begingroup$ Geoff - thanks for the hints, I will do the homework. Actually, this should be simpler, or Serre wouldn't be referring to it in his book (which is supposed to be the introduction to these matters). Well, at least now I know that this is true. BTW, from the examples I looked at, it seems that the order of the composition factors is not fixed: if V, W are factors for P, then for the same P it can be both 0 -> V -> P -> W -> 0 and 0 -> W -> P -> V -> 0, with P still being indecomposable (but this probably should be asked as a separate question). $\endgroup$ – George Dec 10 '12 at 13:16

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