A generator of intersection of two sub-groups. How do I find it? 
For example, an easy one:     $G\:=\:\left(\mathbb{Z},+\right)$ 
H and K are sub-groups: $n\mathbb{Z}, m\mathbb{Z}$ for different $n$ and $m$.
And we know that $n$ and $m$ are generators of their sub-groups.
So I want to find the generators of $n\mathbb{Z}\cap m\mathbb{Z}$
It seems to be: $\dfrac{m\cdot n}{\left(m,n\right)}$ by intuition, am I wrong ? 
And if it is, how do I show that formally? 
 A: If $x\in n\mathbb{Z}\cap m\mathbb{Z}$, then $x$ is an integer multiple of both $m$ and $n$, and therefore an integer multiple of $\text{lcm}(n,m)$. That is, if both $n$ and $m$ divide $x$, then $\text{lcm}(n,m)$ divides $x$. Many proofs of this are given here. This implies that every $x\in n\mathbb{Z}\cap m\mathbb{Z}$ is a multiple of $\text{lcm}(n,m)$, thus:
$$n\mathbb{Z}\cap m\mathbb{Z} = \langle \text{lcm}(n,m)\rangle.$$
So, you were not wrong, since by basic number theory:
$$nm = \gcd(n,m)\text{lcm}(n,m).$$
In general, the intersection of two finitely generated groups is not necessarily finitely generated.
Edit:

We want to show that every $x\in n\mathbb{Z}\cap m\mathbb{Z}$ is an integer multiple of $\text{lcm}(n,m)$ (think of how every $y\in n\mathbb{Z}$ is an integer multiple of $n$, and $n$ generates $n\mathbb{Z}$). 
  If $x\in n\mathbb{Z}\cap m\mathbb{Z}$ then for some integers $k_1,k_2$, 
  $$x=k_1n\quad\text{and}\quad x=k_2m.$$
  By definition, this means that both $n$ and $m$ divide $x$. By the theorem linked above, this implies that $\text{lcm}(n,m)$ divides $x$, which by definition means that there exists a $k\in\mathbb{Z}$ such that 
  $$x=k\cdot\text{lcm}(n,m).$$
  Since $x$ was arbitrary, we can conclude that 
  $$n\mathbb{Z}\cap m\mathbb{Z} = \text{lcm}(n,m)\mathbb{Z}.$$

To see why it has to be the least multiple, consider $3\mathbb{Z}\cap 5\mathbb{Z}$. Notice that $15\in 3\mathbb{Z}\cap 5\mathbb{Z}$, but $15\notin 30\mathbb{Z}$. So if we attempt to make a larger multiple a generator, some elements of $3\mathbb{Z}\cap 5\mathbb{Z}$ will be left out. 
A: The generator of $ mZ \cap nZ $ is the l.c.m of m and n. Intuitively take $3Z \cap 2Z$ it means that you are looking for the smallest common multiple of 2 and 3 which is 6
