Abstract Algebra Homework: Problem Solving on Finite Abelian Groups

This assignment presents a detailed solution to a problem in abstract algebra, specifically focusing on the properties of finite abelian groups. The solution begins by establishing that if a finite abelian group G has a subgroup confined in every subgroup of G, then the order of the subgroup must be prime. The solution then proceeds with a proof by contradiction, analyzing the structure of the group and its subgroups, particularly considering the case where the order of G is divisible by another prime. The solution utilizes concepts such as cyclic groups, maximal subgroups, and the Sylow theorems to demonstrate the properties of the group. The analysis involves examining the subgroups and their intersections, and ultimately demonstrating that the group must be cyclic. Furthermore, the solution explores the implications of the group's structure, including the existence of unique subgroups of specific orders and the relationship between elements of the group. The assignment provides a comprehensive understanding of the group's characteristics and the mathematical reasoning behind the solution.

Abstract/Modern Algebra
Question
Solution
Assuming G is a finite abelian groups with a nontrivial sub-group H confined in every sub-group of G.
Then H should 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 in
every 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 G
1 ≤⟨ h ⟩ ≤⋯ ≤⟨ xp ⟩ ≤ ⟨ x ⟩≤ G .
Since [G:⟨ x ⟩]= p, so y p ∈⟨ x ⟩
Suppose if possible y p is in ⟨ x p ⟩. Then y p=xip for some i. Then considering ( y x−i). its order
is 1∨ p∨p2∨ p3∨… … … … …. .
Subsequently ( y x−i ) p
=1 and y x−i ≠ 1(o . w . y ∈⟨ x ⟩) hence order of ⟨ y x−i ⟩ is p , and it is
equal to the subgroup H=⟨ h ⟩ by hypothesis.
Since y x−1 ∈ H ≤ ⟨ x ⟩, hence y ∈⟨ x ⟩=M contradiction. Thus assumption (that yp ∈⟨ x p ⟩) is
wrong, hence y p is in ⟨ x ⟩ but not in ⟨ x p ⟩: Then ¿ y p∨¿∨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. Having
G=⟨ a ⟩ ⟨ b ⟩ , where b is any element of Gnot in⟨ a ⟩ (where ⟨ a ⟩ denotes the subgroups produced by a).

Quotient by K , the intersections of ⟨ a ⟩∧⟨ b ⟩ , to see that G
K is isomorphic to Zp
z , since G
K is the direct
product of two nontrivial cyclic p−groups (⟨ a ⟩/ K∧⟨ b ⟩ /K ¿and by homomorphism every one of
G/ K ' s subgroups is cyclic. Therefore taking the pre-image of each of the cyclic subgroups of G/ K,
each of which is a distinct maximally proper sub-group of G ,G must have p+1distinct subgroups of
order pn−1 . Also, note that since b was an arbitrary element not in ⟨a⟩ and ⟨ a ⟩/ K∧⟨ b ⟩ /K have the
same number of elements, ⟨ a ⟩ contains every element of order less than pn−1. Since ⟨a⟩ has a unique sub-
group of order pm where n−1>m, Ghas a unique cyclic sub-group of order pm.
.