If $H,K$ are subgroups of $G$ s.t. $o(H), o(K)$ are relativily prime $\implies H \cap K = \{ e \}$.

Question

If $H,K$ are subgroups of $G$ s.t. $o(H), o(K)$ are relativily prime $\implies H \cap K = \{ e \}$.

Here $o(H)$ means the order of $H$.

Below is my attempt but I am afraid I might be jumping some hoops here:

Suppose $a \in H \cap K$ is an arbitrary element. Then we know $o(a)$ must divide $o(H)$ and $o(K)$.

Then $\exists m,n \mathbb{N}$ s.t. $m \cdot o(a) = o(H)$ and $n \cdot o(a) = o(K)$.

But $o(H)$ and $o(K)$ are relatively prime, so for $m \cdot o(a)$ and $n \cdot o(a)$ to be reelatively prime, $o(a)$ must be equal to $1$ which means $a=e$ the identity element.

$\Box$

1. Am I on the right track?
2. Now my problem is with the last statement. How do I effectively argue the last statement?
3. Any alternative proof?

Answer 1

Note, $H\cap K$ is group and moreover subgroup(since, intersection of two subgroup of a same group is again a group ) of both $H, K$. So, $o(H\cap K)|o(H), o(H\cap K)|o(K)$, but since $o(K), o(H)$ are co-primes, so $1$ is the only common divisor to them.

Answer 2

By Lagrange's theorem, the order of $H \cap K$ must divide the orders of both $H$ and $K$. Since the orders of $H$ and $K$ are relatively prime, this means that $H \cap K$ must be trivial.