For an arbitrary n in Z, the cyclic subgroup ⟨n⟩ of Z, generated by n...

Given information:
For an arbitrary n in Z, the cyclic subgroup ${\displaystyle ⟨n⟩}$ 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 ${\displaystyle ⟨m⟩}$ of Z generated by m under addition, and it is the set of multiples of m.
${\displaystyle ⟨m⟩\cap ⟨n⟩}$ is the subgroup that contains multiples of least common multiple of m and n.
Consider the groups generated by ${\displaystyle ⟨m⟩\cap ⟨n⟩}$ and ${\displaystyle ⟨l⟩}$ where l.c.m(m,n) = l.
Let ${\displaystyle x\in ⟨m⟩\cap ⟨n⟩}$
${\displaystyle x\in ⟨m⟩}$ and ${\displaystyle x\in ⟨n⟩}$ then ${\displaystyle x=m{k}_1}$ and ${\displaystyle x=n{k}_2}$ for some $k_1,k_2\in Z$.
Thus, ${\displaystyle m\mid x}$ and ${\displaystyle n\mid x}$ as ${\displaystyle l.c.m.(m,n)=l\Rightarrow l\mid x}$.
Therefore, ${\displaystyle x\in ⟨l⟩}$
Thus, ${\displaystyle ⟨m⟩\cap ⟨n⟩\subseteq ⟨l⟩}$.
Let ${\displaystyle y\in ⟨l⟩}$. then ${\displaystyle y=lk}$.
As l is least common multiple of m and ${\displaystyle n,m\mid l}$ and ${\displaystyle n\mid l}$.
Thus, ${\displaystyle l=ma}$ and ${\displaystyle l=mb}$ for some ${\displaystyle a,b\in Z}$.
${\displaystyle y=ma}$ and ${\displaystyle y=nb}$.
Hence ${\displaystyle y\in ⟨m⟩\cap ⟨n⟩}$.
Thus, ${\displaystyle ⟨l⟩\subset ⟨m⟩\cap ⟨n⟩}$.
Therefore,
${\displaystyle ⟨m⟩\cap ⟨n⟩=⟨l⟩}$.
Therefore, the elements in the subgroup ${\displaystyle ⟨m⟩\cap ⟨n⟩}$ for arbitrary ${\displaystyle m,n\in Z}$ are generated by
${\displaystyle l.c.m(m,n)}$.