$H,K \leq G$, $|H| = 126, |K| = 228$. Show that $H \cap K \leq G$ is either cyclic or isomorphic to $D_3$

Question

So, I don't have to prove that $H \cap K$ is a subgroup of $G$, and $G$ is finite.

I know that any group of order $6$ is isomorphic to either $D_3$ or $\mathbb{Z}_6$ (which is cyclic) so I'm assuming I'm trying to show that $|H \cap K|=6$ and this is perhaps related to gcd(126, 228)=6. However all I'm able to show is:

$|H \cap K |$must divide both $126$ and $228$, the common divisors of $126$ and $228$ are $1,2,3,6$.

If $|H \cap K | = 1 \implies |H \cap K |=\{e\}$ which fits all of the criteria but I'm assuming I'm supposed to be able to show that $|H \cap K | \neq 1,2,3$. But, I just can't figure out how?

Answer 1

You are basically there! $H\cap K$ is a subgroup of order 1, 2, 3, or 6. The only groups of order 1, 2, and 3 are cyclic. This leaves order 6, which is either $\mathbb{Z}/6\mathbb{Z}$ (cyclic) or $D_3$.