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


*

*Am I on the right track?

*Now my problem is with the last statement. How do I effectively argue the last statement?

*Any alternative proof?


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. 
