# 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?

• Every conjugate of a sylow subgroup would be a sylow subgroup. If there is only one, then every conjugate has to coincide. – lulu Jan 3 '18 at 2:34
• Because ALL conjugates have the same order as $P$ so they are still Sylow, but there is only one. – Randall Jan 3 '18 at 2:35
• Thank you lulu for your kind reply. But really I didn't get it . Do you mean it work like conjugacy classes so for any g we have that gPg^_1=P . And also what do you mean when you say they coincide. – Rosa Jan 3 '18 at 2:38
• @Rosa One way to see that $$for any $g \in G$ and Sylow-$p$ subgroup $P$, we have $gPg^{-1}$ is also a Sylow-$p$ subgroup$"$ is to check that the map $\phi: P \to gPg^{-1}$ defined by $a \mapsto gag^{-1}$ is an isomorphism. It then follows that since $P$ is the unique Sylow $p$-subgroup of $G$, we must have $gPg^{-1} = P$. Hence $P$ is normal in $G$. – Alex Vong Jan 3 '18 at 2:55

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 $N_G(P) = G$