# Good explanation on generators of intersection of two cyclic subgroups?

If $\langle a \rangle$ is a cyclic group and $\langle a^n \rangle$ and $\langle a^m \rangle$ are two subgroups then a generator for $\langle a^n \rangle \cap \langle a^m\rangle$ is $a^{\text{lcm}(n,m)}$.

What is an intuitive explanation of why this is so?

I tried to explain it, here are my thoughts:

Say, $S$ is the set of generators of $\langle a^n\rangle$ and $S'$ is the set of generators of $\langle a^m\rangle$ then the generators of $\langle a^n \rangle \cap \langle a^m\rangle$ is $S\cap S'$.

Also, if $a^n$ is a generator then so is $(a^n)^k$ for all $k$ coprime to $|a|$.

So, things in $S\cap S'$ are of the form $a^{kn}$ where $k$ is coprime to $|a|$ and such that $kn = lm$ for some $l$ coprime to $|a|$.

But I don't see how to conclude that $kn = lm$ must be the least common multiple. I mean, obviously it's a multiple, but why must it be the least?

Edit

I guess one can rephrase my question as follows:

Why does $kn = lm =$ least common multiple imply that $k$ and $m$ are coprime to $|a|$?

• I added the tag intuition because I want to gain intuition. I am not looking for a proof of this claim. – a student Jun 6 '16 at 6:17
• Intuition: If $\langle a^n\rangle \cap \langle a^m\rangle = \langle a^i\rangle$ then $n$ should divide $i$ and $m$ should also divide $i$, i.e. $i$ is divisible by $lcm(n,m)$. Since $\langle a^i\rangle$ is intersection of only $\langle a^n\rangle$ and $\langle a^m\rangle$, it is intuitive that $i$ should be $lcm(n,m)$. – p Groups Jun 6 '16 at 8:06

When we prove that every subgroup $H$ of a cyclic group $\langle a\rangle$ is also cyclic, we are looking for the smallest $n$ such that $a^n\in H$ which will become the generator of $H$.
Now for $H=\langle a^m\rangle \cap \langle a^n\rangle$ what is the smallest $i$ such that $a^i\in \langle a^m\rangle$ and also $a^i\in \langle a^n\rangle$?