For an arbitrary n in Z, the cyclic subgroup ⟨n⟩ of Z, generated by n under addition, is the set of all multiples of n. Describe the subgroup ⟨m⟩∩⟨n⟩ for arbitrary m and n ∈ Z.

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Given information: For an arbitrary n in Z, the cyclic subgroup ⟨n⟩ of Z generated by n under addition, and it is the set of multiples of n. Calculation: For an arbitrary n in Z, the cyclic subgroup ⟨m⟩ of Z generated by m under addition, and it is the set of multiples of m. ⟨m⟩∩⟨n⟩ is the subgroup that contains multiples of least common multiple of m and n. Consider the groups generated by ⟨m⟩∩⟨n⟩ and ⟨l⟩ where l.c.m(m,n) = l. Let x∈⟨m⟩∩⟨n⟩ x∈⟨m⟩ and x∈⟨n⟩ then x=mk1 and x=nk2 for some k1,k2∈Z. Thus, m∣x and n∣x as l.c.m.(m,n)=l⇒l∣x. Therefore, x∈⟨l⟩ Thus, ⟨m⟩∩⟨n⟩⊆⟨l⟩. Let y∈⟨l⟩. then y=lk. As l is least common multiple of m and n,m∣l and n∣l. Thus, l=ma and l=mb for some a,b∈Z. y=ma and y=nb. Hence y∈⟨m⟩∩⟨n⟩. Thus, ⟨l⟩⊂⟨m⟩∩⟨n⟩. Therefore, ⟨m⟩∩⟨n⟩=⟨l⟩. Therefore, the elements in the subgroup ⟨m⟩∩⟨n⟩ for arbitrary m,n∈Z are generated by l.c.m(m,n).

For an arbitrary n in Z, the cyclic subgroup \langle n \rangle of Z, generated by n under addition, is the set of all multiples of n. Describe the subgroup \langle m \rangle \cap \langle n \rangle for arbitrary m and n in Z.

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