Let $H$ and $K$ be subgroups of a finite cyclic group $G.$ Prove $|H \cap K| = \gcd(|H|,|K|)$

My attempt:

$H$ and $K$ are subgroups of $G.$ Therefore, $H$ and $K$ are cyclic. Further, $|H|$ and $|K|$ divide $|G|$. Every divisor $m$ of $G$ has a unique cyclic subgroup of order $m$. So

$$H = \langle g^\frac{|G|}{|H|}\rangle, \quad K = \langle g^\frac{|G|}{|K|}\rangle$$

By Lagrange's theorem, $\frac{|G|}{|H|} = [G : H]$ and $\frac{|G|}{|K|} = [G : K]$

$$|H\cap K| = |\langle g^{[G:H]}\rangle \cap \langle g^{[G:K]}\rangle|$$

I don't know how this implies that this equals $\gcd(|H|,|K|)$. Any help would be appreciated.


You have a good start - you need to deal more explicitly with the elements for the critical step, however. In particular, you can write $H = \{g^{n[G:H]} : n\in\mathbb Z\}$ and $K=\{g^{n[G:K]} : n\in\mathbb Z\}$. A reasonable thing to do is to ask which elements are in common between these sets. Note that if you already have a strong handle on sets of the form $n\mathbb Z$, you may be able to very quickly get this result by using that - but if not, you can also do things in a more group-theoretic fashion.

First, observe that $g^{\operatorname{lcm}([G:H],[G:K])}$ is in $H\cap K$ where $\operatorname{lcm}$ is the least common multiple, since the exponent is a multiple of both $[G:H]$ and $[G:K]$ by definition. Let $R=\langle g^{\operatorname{lcm}([G:H],[G:K])}\rangle$ be the subgroup generated by this element. First of all, note that the exponent divides $|G|$ since both $[G:H]$ and $[G:K]$ do, thus the order is just $\frac{|G|}{\operatorname{lcm}([G:H],[G:K])}$. Then you can use some number theory to observe that if $a$ and $b$ divide $n$, then $$\operatorname{lcm}\left(\frac{n}a,\frac{n}b\right)=\frac{n}{\gcd(a,b)}$$ which essentially follows by noting that reciprocation reverses divisibility.

Applying this with $n=|G|$ and $a=|H|$ and $b=|K|$ and using the formulas you gave for the indices of the subgroups gives that $|R| = \gcd(|H|,|K|)$. Then, we are almost done since we know that $R \subseteq H\cap K$. The only remaining step is to note that $R$ is actually all of it - but we know that $|H\cap K|$ divides both $|H|$ and $|K|$, so the intersection cannot have more than $\gcd(|H|,|K|)$ elements, so must be exactly $R$.

  • 3
    $\begingroup$ This is a lot to digest, believe it or not it took me literally all day (around 10 hours or so of studying/solving) to get what I wrote up there. So I'm gonna go to sleep and tackle this with a fresh brain. Thank you for your answer I'll let you know when I start digesting your answer. $\endgroup$ – math_wizard1 Oct 16 '19 at 3:35

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