A question about the consequence of Sylow's theorems The three Sylow theorems are stated here. I don't understand this statement "A consequence of theorem 3 is if $n_p = 1$, then the Sylow p-subgroup is a normal subgroup". 
I understand that if $n_p = 1$, there exists only one Sylow p-subgroup, Say H, and by Theorem 2, $\exists g\in G$ s.t. $gHg^{-1} = H$ but why is it true $\forall g\in G$ so that $H$ becomes normal? Thanks!
 A: Any conjugate $gHg^{-1}$ of any subgroup is itself a subgroup of the same order.  So if $gHg^{-1} \neq H$, we have a second Sylow $p$-subgroup of $G$, contradicting our assumption that there is only one such subgroup.  Thus, $\forall g \in G, gHg^{-1} = H \text{ and } H \triangleleft G.$
A: For each $g\in G$ we have that $gHg^{-1}$ is a group of the same size as $H$, hence it is a $p$-Sylow subgroup as well. Since there is only one $p$-Sylow subgroup we conclude that $gHg^{-1}=H$ for all $g\in G$. This means $H$ is normal in $G$. 
Note that with the same proof we conclude there is a more general fact: if $G$ has only one subgroup of a specific finite size $d$ then this subgroup must be normal in $G$. The converse statement (that if $H\trianglelefteq G$ then there are no other subgroups of the same size as $H$) is obviously false in general. However, for Sylow subgroups the converse is also true and it follows from the second Sylow theorem. 
A: From the theorem, $n_p=1$ implies
$$\lvert G : N_G(P)\rvert = 1,$$
where $N_G$ denotes the normalizer. This tells you that $N_G(P)$ is all of $G$ (in particular,  since we're working with finite groups, $ \lvert G : N_G(P)\rvert = \frac{\lvert G \rvert}{\lvert N_G(P)\rvert}$).
