# My argument on normalizer of Sylow-P subgroups

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?

• $gPg^{-1}$ can be $P$ itself, just think about the case where the $p$-sylow is normal. – AnalysisStudent0414 Nov 23 '16 at 19:04
• Yes, but if $P$ is normal then the group has only $1$ Sylow-p subgroup, right? – Ninja Nov 23 '16 at 19:06
• Yes, but I meant that your deduction "if $g$ is not in $P$, $P^g$ is a different subgroup" doesn't really make sense. – AnalysisStudent0414 Nov 23 '16 at 19:11
• The number of Sylow $p$-subgroups is $|G:N(P)|$, so certainly $N(P)$ is not always $P$ itself. E.g. if there is only one Sylow $p$-subgroup, then it is characteristic, hence normal, so its normalizer is all of $G$. At the other extreme, if $N(P) = P$ then there must be $|G:P|$ Sylow $p$-subgroups. This isn't always possible since the number of Sylow $p$-subgroups must also be congruent to $1$ mod $p$. – Bungo Nov 23 '16 at 19:17

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.

• I see. Then how can we assure that $gPg^{-1}$ is another Sylow-p subgroup? – Ninja Nov 23 '16 at 19:10
• @Ninja $gPg^{-1}$ is isomorphic to $P$ (conjugation is an automorphism), so they have the same size. – Bungo Nov 23 '16 at 19:12
• Okay, my final question is assume that we have a finite group $G$ and Sylow-p-subgroups $P_1, P_2, \dots, P_n$ and for $g_2, g_3, \dots, g_n \in G: g_2P_1g_2^{-1} =P_2, \dots, g_k P_1 g_k^{-1} = P_k, \dots, g_nP_1 g_n ^{-1} = P_n$. I saw that even if $g \notin P, gPg^{-1} = P$. So is there a partition of $G$ such that there are at least $n$-many classes and $n-1$-many class' representatives are $g_2, g_3, \dots g_n$ for the Sylow-p subgroups? – Ninja Nov 23 '16 at 19:23
• Bungo's comment to the original question is a good answer I think. $G$ acts by conjugation on its set of Sylow p-subgroups. This action is transitive --- if $P$ and $Q$ are any two Sylow p-subgroups, there must exist some $g$ in $G$ such that $gPg^{-1}=Q$. Therefore the number of Sylow p-subgroups is $[G:N(P)]$. It can also be shown that the number of Sylow p-subgroups is always 1 mod p. – DCarter Nov 23 '16 at 19:30
• @Ninja When you say "is another Sylow-$p$ subgroup" you need to clarify whether you mean "is also Sylow-$p$" or "is a Sylow-$p$ subgroup distinct from $P$". The English word "another" is too vague to specify your intent clearly. – Erick Wong Nov 23 '16 at 20:47

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 $5$.