Multiplicity free subgroups of semi-direct products
نویسندگان
چکیده
منابع مشابه
Two subgroups and semi-direct products
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: Indagationes Mathematicae
سال: 2009
ISSN: 0019-3577
DOI: 10.1016/s0019-3577(09)80002-5