Generator of intersection of two cyclic groups 
Let $H= \langle n \rangle$ and $K= \langle m \rangle$ be two cyclic groups.  Show that their intersection is a  cyclic subgroup generated by the lcm of $n$ and $m$.

I took an element, say $a$, belonging to $H \cap K$.  Then $a$ can be written as a multiple of $m$ and as a multiple of $n$.  Then I want to show that it can be written as a multiple of lcm of $n$ and $m$.
 A: Your approach is good.  You're trying to show $H \cap K = \langle \text{lcm}(m,n) \rangle$.  To show this, first prove $H \cap K \subseteq \langle \text{lcm}(m,n) \rangle$; then prove $H \cap K \supseteq \langle \text{lcm}(m,n) \rangle$.
To show $H \cap K \subseteq \langle \text{lcm}(m,n) \rangle$, you're doing the right thing.  Take any element of $H \cap K$ and call it $a$.  Since $a$ is a multiple of both $m$ and $n$, it's a multiple of $\text{lcm} (m,n)$.  (If you're not allowed to state this without proof, use unique prime factorization.)  Hence $a \in \langle \text{lcm}(m,n) \rangle$.
Now you just need to show $H \cap K \supseteq \langle \text{lcm}(m,n) \rangle$.
Go the other way around - take some element of $\text{lcm}(m,n)$, call it $b$.
You know that $b$ is a multiple of $\text{lcm} (m,n)$; you just need to show it's in both $H$ and $K$.  I trust you can do this.  Then you'll be done - since $H \cap K \subseteq \langle \text{lcm}(m,n) \rangle$ and $H \cap K \supseteq \langle \text{lcm}(m,n) \rangle$, we must have $H \cap K = \langle \text{lcm}(m,n) \rangle$.
