Conference on Group Theory, University of by [EDS] R.W. GATTERDAM AND K.W.WESTON


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Then A ≤ ζ(G). Proof. Suppose the contrary, that is ζ(G) contains no A. Then Aζ(G)/ζ(G) is a non-identity normal subgroup of the hypercentral group G/ζ(G), and hence there exists some a ∈ (A ∩ ζ2 (G)) \ ζ(G). Since A is a p-subgroup, we may assume that ap ∈ ζ(G). Then the mapping φ : g → [g, a], g ∈ G is an endomorphism of G such that [G, a] = Im φ = 1 since a ∈ ζ(G). Actually, since ap ∈ ζ(G), [G, a]p = 1 , and so Im φ is an elementary abelian p-subgroup. Since [G, a] = Im φ ∼ = G/ Ker φ = G/CG (a), we deduce that G/CG (a) is an elementary abelian p-group as well.

3. Let R be a ring, and let A be an R-module. Then the following statements are equivalent. (1) A is a sum of simple R-submodules. (2) A is a direct sum of simple R-submodules. (3) For every R-submodule B, there is an R-submodule C such that A = B ⊕ C. Proof. 1. (2) ⇒ (3). Let B be a non-zero R-submodule of A. 2, B = λ∈Λ Mλ , where Mλ is a simple R-submodule for each λ ∈ Λ. This means that the family {Mλ | λ ∈ Λ} is independent. Then there is a maximal independent family M, such that {Mλ | λ ∈ Λ} ⊆ M.

Since H = K∈L K and aF H = K∈L aF K, we deduce that aF H is further a simple F H-submodule. 4. If A is a simple ZG-module, then the underlying additive group of A either is a divisible torsion-free abelian group or is a p-elementary abelian group, for some prime p. In the first case we may think of A as a QG-module whereas in the second one we think of A as an Fp G-module. 7 the following result. 8. Let G be an abelian group and H a periodic subgroup of G. If A is a simple ZG-module, then A is a semisimple ZG-module.

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