My argument on normalizer of Sylow-P subgroups

Question

Given a group $G$ and a Sylow-p subgroup $P$ of $G$. If necessary -I am not sure whether it is or not- assume that $G$ has more than one Sylow p-subgroups. I am claiming that normalizer $N(P)$ of $P$ is itself. My arguement is the following:

If $P \subset N(P)$, then there is an element $g$ in $N(P)$ that is not in $P$. Since $g$ is an element of $N(P)$, we have $gPg^{-1}=P$. However, we know that $gPg^{-1} = Q$ is another Sylow-p subgroup of $G$. Then we get $P=Q$ which makes no sense so $N(P)=P$.

Is my argument correct?

Answer 1

Your argument is incorrect. It can happen that a Sylow p-subgroup $P$ of $G$ is also a normal subgroup of $G$. The Sylow 3-subgroup of the nonabelian group of order 6 is normal, for example. If $P$ is a normal subgroup of $G$ then $N(P)=G$.

You are right to say that $gPg^{-1}$ is a Sylow p-subgroup of $G$ whenever $P$ is such, but it's possible that $P = gPg^{-1}$ even when $g$ is not in $P$.

I see that I need to add an example in which the Sylow p-subgroup is not unique. The nonabelian group $S_3$ of order 6 has exactly 3 Sylow 2-subgroups. Suppose $G$ is the direct product of $S_3$ and a cyclic group of order 5. Then $G$ still has three Sylow 2-subgroups. But the normalizer of each of these has order 10 -- all the elements of order 5 in G are in the normalizer of each Sylow 2-subgroup.

Answer 2

The argument is incorrect because even if $g \not\in P$, $P^g = gPg^{-1}$ can be equal to $P$, and this can happen even if $G$ has more than one $p$-Sylow.

What is true is that for all $P,Q$ $p$-Sylow subgroups of $G$, there exists one element $g$ such that $P^g = Q$ (usually known as Sylow's third theorem)

If you want an explicit example, in $A_5$ the $2$-Sylow has order $4$, so for example can be generated by $(12)(34), (13)(24)$. Conjugation by $(134)$ leaves the group fixed, since this subgroup is actually normal in $A_4<A_5$. Try to compute it explicitly if you are not convinced by the argument.

Obviously $A_5$ has more than one $2$-Sylow, obtain another just by conjugation with any element which does not fix $5$.