Given information:

For an arbitrary n in Z, the cyclic subgroup $\u27e8n\u27e9$ 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 $\u27e8m\u27e9$ of Z generated by m under addition, and it is the set of multiples of m.

$\u27e8m\u27e9\cap \u27e8n\u27e9$ is the subgroup that contains multiples of least common multiple of m and n.

Consider the groups generated by $\u27e8m\u27e9\cap \u27e8n\u27e9$ and $\u27e8l\u27e9$ where l.c.m(m,n) = l.

Let $x\in \u27e8m\u27e9\cap \u27e8n\u27e9$

$x\in \u27e8m\u27e9$ and $x\in \u27e8n\u27e9$ 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 $l.c.m.(m,n)=l\Rightarrow l\mid x$.

Therefore, $x\in \u27e8l\u27e9$

Thus, $\u27e8m\u27e9\cap \u27e8n\u27e9\subseteq \u27e8l\u27e9$.

Let $y\in \u27e8l\u27e9$. 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 $y\in \u27e8m\u27e9\cap \u27e8n\u27e9$.

Thus, $\u27e8l\u27e9\subset \u27e8m\u27e9\cap \u27e8n\u27e9$.

Therefore,

$\u27e8m\u27e9\cap \u27e8n\u27e9=\u27e8l\u27e9$.

Therefore, the elements in the subgroup $\u27e8m\u27e9\cap \u27e8n\u27e9$ for arbitrary $m,n\in Z$ are generated by

$l.c.m(m,n)$.