 Armorikam

2021-08-04

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 $⟨m⟩\cap ⟨n⟩$ for arbitrary m and n $\in$ Z. bahaistag

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.
$⟨m⟩\cap ⟨n⟩$ is the subgroup that contains multiples of least common multiple of m and n.
Consider the groups generated by $⟨m⟩\cap ⟨n⟩$ and $⟨l⟩$ where l.c.m(m,n) = l.
Let $x\in ⟨m⟩\cap ⟨n⟩$
$x\in ⟨m⟩$ and $x\in ⟨n⟩$ then $x=m{k}_{1}$ and $x=n{k}_{2}$ for some ${k}_{1},{k}_{2}\in Z$.
Thus, $m\mid x$ and $n\mid x$ as $l.c.m.\left(m,n\right)=l⇒l\mid x$.
Therefore, $x\in ⟨l⟩$
Thus, $⟨m⟩\cap ⟨n⟩\subseteq ⟨l⟩$.
Let $y\in ⟨l⟩$. then $y=lk$.
As l is least common multiple of m and $n,m\mid l$ and $n\mid l$.
Thus, $l=ma$ and $l=mb$ for some $a,b\in Z$.
$y=ma$ and $y=nb$.
Hence $y\in ⟨m⟩\cap ⟨n⟩$.
Thus, $⟨l⟩\subset ⟨m⟩\cap ⟨n⟩$.
Therefore,
$⟨m⟩\cap ⟨n⟩=⟨l⟩$.
Therefore, the elements in the subgroup $⟨m⟩\cap ⟨n⟩$ for arbitrary $m,n\in Z$ are generated by
$l.c.m\left(m,n\right)$.

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