Finding order of group intersection

Question

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 $|H_1 \cap H_2|=5$

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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!

Answer 1

$\;G\;$ 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...?)

Answer 2

When $G$ is cyclic, of finite order $n$, the lattice of subgroups of $G$ is isomorphic to the lattice of the positive divisors of $n$, so a subgroup of order 15 and a subgroup of order 25 will intersect in a subgroup of order $\gcd(15, 25) = 5$.

Another answer has already shown that this does not necessarily work anymore if $G$ is not cyclic.

Answer 3

For non-cyclic $G$, you can't solve it unless you know something more about the subgroups than just their orders. For instance, if $G = \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$, then $G$ has two subgroups of order 2 with trivial intersection.