Proving $o(H\cap K)=\gcd(o(H),o(K))$.

Question

Let $H$ and $K$ be subgroups of a finite cyclic group $G,$ then $|H \cap K| = \gcd(|H|,|K|)$.
*The same question is already asked in this plateform here.

So this is indeed a duplicate question but i am not satisfied by the answer.

Interestingly that question was also duplicate but i did not find the answer in the given link.

Although it was a good answer but i think there must be a simple proof for this statement.
My attempt:Since $o(H\cap K)\:\text {divides o(H) and o(K)}$.
Therefore, $o(H\cap K)\:\text{divides gcd(o(H),o(K))}$. Now if we are able to prove that $\text{gcd(o(H),o(K)) divides}\:o(H\cap K)$,then we are done.
Kindly help me.Thank you.

Answer 1

One way of doing this is to use the isomorphism with $\mathbb{Z}/N\mathbb{Z}$ for some $N$.

A direct way is to realize that you have not yet used the fact that $G$ is cyclic. So let the order of $G$ be $N$ and let $g$ be a generator. Then $G=\{g^j:0\leq j\leq N-1\}$.

What does a subgroup $H$ of $G$ look like? Well say some power $g^a$ is in $H$. Then $g^{an}\in H$ for all $n$. So if $a$ is coprime with $N$, then $g\in H$ and $H=G$. Therefore, you only get proper subgroups if they are generated by some $g^a$ with $a$ being a divisor of $N$.

Therefore, the only interesting case is when $H$ and $K$ are proper subgroups of $G$. Let $g^a$ and $g^b$ be their respective generators.

Then, $|H|=\tfrac{n}{a}$ and $|K|=\tfrac{n}{b}$. The gcd of these is clearly $\tfrac{n}{\text{lcm}(ab)}$. But the group $H\cap K$ is generated by $g^{\text{lcm}(ab)}$, so the claim follows.