A problem of subgroup of a group If $G$ be a finite group of order $pq$ where $p$ & $q$ are prime numbers such that $p>q$. Prove that $G$ has at most one subgroup of order $p$.Hence prove that a group of order 6 has at most one subgroup of order 3.
My try is that i let $G$ has two subgroup $H$ & $K$ of same order $p$.Then we have to prove $H=K$
since $p>q$ & $o(G)=pq$ & clearly $p>\sqrt{pq}$
hence $o(H)>\sqrt{o(G)}$ & also $o(K)>\sqrt{o(G)}$
so there must some elements in intersection of both of the subgroups with identity i.e
$H\cap K$$\neq ${e}
now since o($H\cap K$) must divide $o(H)$ & $o(G)$.And from above conclusion $o(H\cap K)\neq 1$ & $p$ is prime so $o(H\cap K)$ must be $p$ hence
$H=K$
but i am not sure about the line which is 'bold'.
plz suggest me.

This is correct, though you have not explained that.
To do that, consider the map $H\times K\to G$ defined as $(h,k)\mapsto hk$. What can you say about it if $H\cap K=\{e\}$?
