Resolutions by permutation modules
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Archiv der Mathematik
Resolutions by permutation modules Paul Balmer and Dave Benson
Abstract. We prove that, up to adding a complement, every modular representation of a finite group admits a finite resolution by permutation modules. Mathematics Subject Classification. 20C20. Keywords. Modular representation theory, Permutation modules, Finite resolution, Elementary abelian subgroup.
Let G be a finite group and k be a field of characteristic p > 0 dividing the order of G. It is well-known that if G has non-cyclic Sylow p-subgroups, the k-linear representation theory of G is complicated. In particular, the Krull– Schmidt abelian category, kG-mod, of finite-dimensional kG-modules admits infinitely many isomorphism classes of indecomposable objects. On the other hand, there is a much simpler class of kG-modules, the permutation modules, i.e., those isomorphic to kX for X a finite G-set. The finite collection {k(G/H)}HG additively generates all such modules. For a kG-module M ∈ kG-mod, we want to analyze the existence of what we will call a permutation resolution for short, i.e., an exact sequence 0 → Pn → Pn−1 → · · · → P1 → P0 → M → 0
(1)
where all Pi are permutation modules. Up to direct summands, the following is always possible: Theorem 2. Let G be a finite group and M ∈ kG-mod. Then there exists a kG-module N such that M ⊕ N admits a finite resolution (1) by permutation modules. The related problem of resolutions (1) that are not only exact but remain exact under all fixed-point functors has been recently discussed in [2]. Allowing p-permutation modules Pi (that is, direct summands of permutation modules), Paul Balmer supported by NSF Grant DMS-1901696.
P. Balmer and D. Benson
Arch. Math.
Bouc–Stancu–Webb prove that such resolutions exist for all M if and only if G has a Sylow subgroup that is either cyclic or dihedral (for p = 2). Unsurprisingly, Theorem 2 reduces to a Sylow subgroup S of G since every G G M is a direct summand of IndG S ResS (M ) and since the functor IndS is exact and preserves permutation modules. So we focus on the case where G is a p-group. For the proof, we shall consider a stronger property: Definition 3. We say that a resolution (1) is free up to degree m 0 if Pi is a free module for i = 0, . . . , m. We say that M admits good permutation resolutions if for every integer m 0, there exists a finite resolution (1) by permutation modules that is free up to degree m. Remark 4. Let G be a p-group. A kG-module M admits good permutation resolutions if and only if for all m 1, the mth Heller loop Ωm M admits a finite permutation resolution. Also, if Q is free and M ⊕ Q admits a permutation resolution as in (1), then the epimorphism P0 M ⊕ Q Q forces Q to = be a direct summand of P0 and one can remove 0 → Q − → Q → 0 from the resolution. So, if M ⊕ Q has a permutation resolution that is free up to degree m, then so does M . An advantage of good permutation resolutions is the two out of three property: Proposition 5. Let G be a p-group. Let 0 → L → M → N → 0 be an exact sequence of k
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