How many elements are there in the intersection of two subgroups of a finite cyclic group? Let's assume that we have two subgroups $H_1$ and $H_2$ in $\mathbb{Z}_n$ with $k$ and $l$ elements respectively. 
How many elements are there in the intersection $H_1\cap H_2$? 
Let denote this by $m$. By Lagrange's theorem we have:
$ m|\gcd(k,l)$. Maybe $m=\gcd(k,l)$ ?
 A: Yes, you are half-way there.
Let $C_n=\langle a\rangle$ be a cyclic group generated by an order $n$ element $a$.
Key fact The mapping $m\longmapsto C_m=\langle a^{n/m}\rangle$ is a bijection from the set of all positive divisors of $n$ onto the set of all subgroups of $C_n$. Note that $C_m$ is precisely the order $m$ subgroup of $C_n$.
What you are trying to prove follows from the following.
Consequence For any two positive divisors $k$ and $l$ of $n$, we have $C_k\cap C_l=C_{\gcd(k,l)}$.

Proof  Let $g:=\gcd(k,l)$ and write $k=k_0g$, $l=l_0g$. Then $n/g=k_0 \cdot (n/k)=l_0\cdot (n/l)$ whence $a^{n/g}$  lies in $C_k\cap C_l$ and therefore $C_g\subseteq C_k\cap C_l$. Now as you observed, Lagrange entails that $|C_k\cap C_l|\leq g =|C_g|$. So the result follows by cardinality. $\Box$

A: Note that in any finite cyclic group $G$, there is exactly one subgroup of order $d$ whenever $d$ divides $|G|$, and it consists precisely of the elements of order $d$. Therefore in the cyclic group $\Bbb Z/n\Bbb Z$, the subgroups $H_1$ and $H_2$ are those of the elements of order $k$ and $l$, respectively. The order of the elements of $H_1\cap H_2$ is a common divisor of $k$ and $l$, hence divides $d=\gcd(k,l)$. Since this is a necessary and sufficient condition, $H_1\cap H_2$ consists of all $d$ elements of order $d=\gcd(k,l)$ in $\Bbb Z/n\Bbb Z$.
