Let $M = p\mathbb{Z}$ and $N = q\mathbb{Z}$ be two subgroups of $(\mathbb{Z}, +)$, show that $M \cap N = lcm(p, q)\mathbb{Z}$ 
Let $M = p\mathbb{Z}$ and $N = q\mathbb{Z}$ be two subgroups of $(\mathbb{Z}, +)$, where $p, q > 0$. Show that $M \cap N = l\mathbb{Z}$ where $l = lcm(p, q)$.

It's not too hard to show that $l\mathbb{Z} \subseteq M \cap N.$ Choose $x \in l\mathbb{Z}$, then $x = lz$ for some $z \in \mathbb{Z}$. Now since $l = pa$ for some $a \in \mathbb{Z}$ (since $l$ is a multiple of $p$), it follows that $x =paz \in p \mathbb{Z} = M$. A similar argument shows that $x \in q\mathbb{Z} = N$, hence $x \in M \cap N$.
Now I'm finding some difficulty in showing that $M \cap N \subseteq l\mathbb{Z}$. My idea was this, choose $y \in M \cap N$. Then $y = pz$ and $z = qw$ for some $z, w \in \mathbb{Z}$. By the division theorem for the integers we get $$z= l \gamma + r \text{ and } w = l\theta +r'$$ where $\gamma, \theta \in \mathbb{Z}$ and $0 \leq r < \gamma$ and $0 \leq r' < \theta$. Then we have $y = l(p\gamma) + pr$ and $y = l(q\theta) + qr'$...

Now this is the point where I got stuck, I was thinking that the next thing to do would be to show $r = r' = 0$ and then  complete the proof that way, however I'm not really sure how to go about doing that. 
 A: $y\in M\cap N$ means that $y=pz$ and $y=qw$ for some $z,w\in \mathbb{Z}.$ Furthermore, $l=px$ and $l=qy$ for some $x,y\in \mathbb{Z}.$ Now say that 
$$y = lm+n$$ by the remainder theorem with $0<n<l$ (assume that $n\neq 0$ to get a contradiction) and $m\in \mathbb{Z}$. Then
$$pz = pxm+n\implies p(z-xm)=n.$$
So $p|n.$ Similarily $q|n.$ Since $l$ is the least common multiple it must be that $n\geq l$ which is not possible and so $n=0$ as desired. 
A: If z be an element in the intersection of M and N, then z is a multiple of both p and q. Thus z= pqr for some r belonging to integers. Being a group under addition, it contains all the  multiples of pq. Now if the set contain any other element, then we can write such y as y= pqk +r where r is strictly less than pq and greater than or equal to 0. From this, we deduce, that y- pqk = r and since both y and pqk belongs to this set, r also belong to this set. But then r is 0. Because no element less than pq and greater than 0 is divisible by pq.
A: The lowest common multiple $l$ of $p$ and $q$ satisfies


*

*$p\mid l$ and $q\mid l$

*for all $m$, if $p\mid m$ and $q\mid m$, then $l\mid m$.


It's easy to prove that such a number is unique, if we only allow nonnegative integers.

A well-known and very important fact about subgroups of $\mathbb{Z}$ is that each of them can be written as $n\mathbb{Z}$ for a unique integer $n\ge0$.

Thus you know that $p\mathbb{Z}\cap q\mathbb{Z}=l\mathbb{Z}$ for a unique $l\ge0$. Clearly, $l\mathbb{Z}\subseteq p\mathbb{Z}$, so $p\mid l$. Similarly $q\mid l$.
Now suppose $m$ is a common multiple of $p$ and $q$. This implies $m\mathbb{Z}\subseteq p\mathbb{Z}$ and $m\mathbb{Z}\subseteq q\mathbb{Z}$. Hence
$$
m\mathbb{Z}\subseteq p\mathbb{Z}\cap q\mathbb{Z}=l\mathbb{Z}
$$
and therefore $l\mid m$.
We proved $l$ satisfies the properties 1 and 2 of the lowest common multiple, so $l=\operatorname{lcm}(p,q)$.

In a fairly similar fashion you can prove that $p\mathbb{Z}+q\mathbb{Z}=d\mathbb{Z}$ where $d=\gcd(p,q)$ (because $p\mathbb{Z}+q\mathbb{Z}$ is the smallest subgroup containing both $p\mathbb{Z}$ and $q\mathbb{Z}$, dual to the fact that $p\mathbb{Z}\cap q\mathbb{Z}$ is the largest subgroup contained in the two subgroups).
