I just checked many many questions considering this subject but couldn't find a general answer.
If we're trying to prove that a group G is not simple and we're counting elements. In my book it says that we cannot in general (in particular if we have p-Sylows with order >1) assume the intersection of all p-Sylows for given p (say $\bigcap_iP_i$ for $i\in \{1,...,s_p(G)\}$) to be trivial. I understand that we cannot do this in general, but if the intersection wasn't trivial, then $\bigcap_i P_i$ was a normal subgroup of G. How can G be simple then?
The unique largest normal $p$-subgroup of a finite group $G$ is $O_p(G)=\cap_{P \in Syl_p(G)}P$. This subgroup does not have to be trivial, depending for example on the prime $p$: $O_2(S_3)=1$, but $O_3(S_3)=A_3$. For a non-abelian simple group, irrespective of the prime $p$, $O_p(G)=1$.