Suppose $H$ is a finite group. Let $p,q$ be two primes and $S_q,S_p$ the associated Sylow $p$-subgroups of $H$.
If $p\ne q$, then $S_q\cap S_p=\{e\}$.
Suppose $p\ne q$. Let $|S_p|=p^r$ and $|S_q|=q^k$ for some $r,k\in\mathbb{N}$.
Since $S_q,S_p\leq H$, $S_q\cap S_p\leq H$. That shows that $e\in S_q\cap S_p$.
Let $x\in S_q\cap S_p$: then $x\in S_q$ and $x\in S_p$. This implies that $x^{p^r}=e=x^{q^k}$. Thus $x^{p^r}=x^{q^k}$. How can I use this latter fact together with $p\ne q$ to conclude that $x=e$?
I am not allowed to use generalized euclid's lemma, GCD, LCM, "relatively prime", etc.
By Lagrange's theorem, the order divides $p^i$ and $q^j$. But $(p^i,q^j)=1$.