Finding order of group intersection Let $G$ be cyclic group, and $H_1, H_2$ subgroups. $|H_1|=15$, and $|H_2|=25$ Find $|H_1 \cap H_2|$.
So this is the solution we were presented at recital:
$|H_1|$ and $|H_2|$ divides $|G|$, so $|G|=lcm(15,25)=75k$, $k \in \mathbb N$. Since $G$ is cyclic, so are its subgroups, so $H_1=<\frac {|G|}{|H_1|}>=<\frac {75k}{15}>=<5k>$. Same with $H_2$ we get $H_2=<3k>$.
Then $H_1 \cap H_2=<lcm(5k,3k)>=<15k>$, and then we get $H_1 \cap H_2=<\frac{|G|}{|H_1 \cap H_2|}> \Rightarrow <15k>=<\frac{75k}{|H_1 \cap H_2|}> $, hence $|H_1 \cap H_2|=5$

But with a quick observation one can see that $|H_2 \cap H_2|$ divides both $|H_1|$ and $|H_2|$, so one can say $|H_2 \cap H_2|=gcd(|H_1|,|H_2|)=5$. 
Is my solution correct? Is it correct always, or only when $G$ is cyclic? 
And how could you then solve the question for non-cyclic $G$?
Thanks!  

$\;G\;$ cyclic and finite (why?) , so it has one unique subgroup of each order dividing  its order.
$$|H_1\cap H_2|\;\mid \;15\,,\,25\implies |H_1\cap H_2|=1,5$$
But there's a subgroup of $\;G\;$ of order $\;5\;$, and since any subgroup of a cyclic one is cyclic, this subgroup of order $\;5\;$ is a subgroup both of $\;H_1\;$ and of $\;H_2\;$, and we're done.
In case $\;G\;$ is not cyclic you cannot prove this (counterexample...?)
