Q:

Sylow normal subgroups

My question is regarding normality of sylow subgroup. In the second sylow subgroup we know that $p$-sylow subgroups conjugate. And I noticed that when we trying to prove simplicity of a group, whenever we found that the number of $p$-sylow subgroups is $1$ we conclude that it is normal and hence the group is not simple. However the second theorem of sylow said that ( there exist $g$ such that $gPg^{-1} = P$) but in a normal subgroup we have to have all $g$ of $G$ satisfying this not just one $g$.
Can any body explain why we say that $p$-sylow subgroup is normal when we have only one of them? 

A:

The proof of Sylow's theorems is done by letting the group act on the Sylow $p$-subgroups through conjugation.  (Or at least one proof - the one I know of - is constructed through this way)
Let $G$ be the proof and let $\mathrm{Syl}_p(G)$ be the set of Sylow $p$-subgroups. Let $P\in \mathrm{Syl}_p(G)$. If the number of Sylow $p$-groups is 1, this means that the orbit of $P$ is just 1 length. This implies $P^G = P$. Or in other words: $(\forall g\in G)(P^g=P)$.
You can also see this through the third Sylow Theorem:
$$
n_p(G) = |G: N_G(P)|
$$
If $n_p(G)=1$ then $|N_G(P)| = |G|$. Now $N_G(P)$ is by definition: $|\{g\in G: P^g = P\}|$. If this should have the same order as $G$ then $N_G(P) = G$

