Show or give a counterexample: $K\cap H$ is cyclic.

Question

Show or give a counterexample: Let $G $ be a group with $|G| = 3\cdot 5 \cdot 7$ and K and H are subgroups of G. Suppose that H is not a subgroup of K and vice versa. Now $K\cap H$ is cyclic.

I can't really come up with a counter example and I am stuck at the proof. If i can proof that $|K\cap H|$ is prime then I know that the group is cyclic and 3, 5 and 7 already are prime.

Answer 1

Hint neither $H$ not $K$ can be $G$. Then $|H|$ and $|K|$ are distinct proper divisors of $3 \cdot 5 \cdot 7$, and hence divisible by at most 2 primes.

Show that $|H \cap K|$ cannot be divisible by 2 distinct primes.

Answer 2

Hint: Neither $K$ nor $H$ is the whole group, so their intersection is either trivial or has prime order.

Answer 3

As $H\not\subset K$ and $K\not\subset H$, both $H$ and $K$ are proper subgroups of $G$, and $H\cap K$ is a proper subgroup of both $H$ and $K$. Hence $|H|$ and $|K|$ are both proper divisors of $|G|=3\times5\times7$, and $|H\cap K|$ is a proper divisor of both $|H$| and $|K|$. It follows that $|H\cap K|$ is prime or equal to $1$, so $H\cap K$ is cyclic.