Modern Algebra Assignment
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Added on 2020-05-11
Modern Algebra Assignment
Added on 2020-05-11
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Abstract/Modern AlgebraQuestionSolution Assuming G is a finite abelian groups with a nontrivial sub-group H confined in every sub-group of G.Then Hshould be of prime order (otherwise it will have a proper subgroup K and H⊈K).Assuming |H|=p. If |G| is divisible by another prime q, then there will be a subgroup of order q and H can not be contained in it. Hence ¿G∨¿pk for some k, with condition that there exist a sub-group H of order p confined inevery subgroup.Assuming H=⟨h⟩. If k=1 we are done. Letk>1. Then each proper sub-groups of G satisfy the stated hypothesis and so is cyclic by induction.Assume M is maximal sub-group of G; it is cyclic, say M=⟨x⟩. Take y∈G∖M Then by considering the following situation in G1≤⟨h⟩≤⋯≤⟨xp⟩≤⟨x⟩≤G.Since [G:⟨x⟩]=p, so yp∈⟨x⟩Suppose if possible yp is in ⟨xp⟩. Then yp=xip for some i. Then considering (yx−i). its order is 1∨p∨p2∨p3∨.................Subsequently (yx−i)p=1 and yx−i≠1(o.w.y∈⟨x⟩) hence order of ⟨yx−i⟩ is p, and it is equal to the subgroup H=⟨h⟩ by hypothesis.Sinceyx−1∈H≤⟨x⟩, hence y∈⟨x⟩=M contradiction. Thus assumption (thatyp∈⟨xp⟩) is wrong, henceyp is in ⟨x⟩ but not in ⟨xp⟩: Then ¿yp∨¿∨x∨so∨y∨¿p.∨x∨¿∨G∨.Setting ¿G∨¿pn and by an induction onn. It is clear wheren=1. Then, by induction each proper subgroups of Gare cyclic. By the Sylow theorems G therefore consists elements of order pn−1. HavingG=⟨a⟩⟨b⟩, where b is any element of Gnot in⟨a⟩ (where ⟨a⟩ denotes the subgroups produced by a).
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