I would like to investigate a few properties of Sylow p-subgroup, especially, when $p^n \;|\; |G|$ for $n > 1$.
If $n = 1$ and $P_1, P_2$ are Sylow $p$-subgroups, then if $id \neq g \in P_1 \cap P_2$, then $\{g, \cdots, g^p\} = P_1 = P_2$, so the proof is done. I would like to know what happens if $n \geq 2$; I cannot use the same trick as before since there's is no guarantee that a Sylow p-subgroup $P \cong \mathbb Z/p^n$
(I am aware of the result that if $P_1, P_2$ are Sylow $p$-subgroup, then $P_1$ and $P_2$ are conjugate to each other. If I were to use it, then it is sufficient to show that if $H_1, H_2 < G$ such that $H_1, H_2$ are conjugates, then $H_1 \cap H_2 = \{id\}$ or $H_1 = H_2$. But how, and is it even true? So really, I am asking two separate questions here.)
Best Answer
No, there's no reason this should be true. A simple conceptual reason is that the center $Z(G)$ may have nontrivial elements of order $p$, in which case the Sylow $p$-subgroup of $Z(G)$ is a subgroup of every Sylow $p$-subgroup. So we can consider, for example, the Sylow $2$-subgroups of $C_2 \times S_3$, which has center $C_2$, as ancientmathematician says in the comments. More generally the same argument works if $G$ has a normal $p$-subgroup.