Free resolutions for semi-direct products

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Crossed complexes, free crossed resolutions and graph products of groups

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Proof. Clearly, the image of F is HK by definition. To see that F is injective, suppose that F (h1, k1) = F (h2, k2). Then by definition h1k1 = h2k2. Thus h −1 2 h1 = k2k −1 1 . Since H is a subgroup, h−1 2 h1 ∈ H, and since K is a subgroup, k2k −1 1 ∈ K. Thus h−1 2 h1 = k2k −1 1 ∈ H ∩ K = {1}, and so h −1 2 h1 = k2k −1 1 = 1. It follows that h−1 2 h1 = 1, so that h1 = h2, and similarly k2k −1 ...

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ژورنال

عنوان ژورنال: Tohoku Mathematical Journal

سال: 1993

ISSN: 0040-8735

DOI: 10.2748/tmj/1178225847