# T/F questions on $p$-Sylow subgroups and normalizer

Prove the following statement if it is true. Otherwise, disprove it by giving a counterexample.

1) The normalizer $N$ in a finite group $G$ of a subgroup $H$ of $G$ is always a normal subgroup of $G$.

2) A $p$-Sylow subgroup of a finite group $G$ is normal in $G$ iff it is the only $p$-Sylow subgroup of $G$.

I try to prove both of them. Please take a look at my answers. If they are indeed true, please help me by answering my question below. If they are false, I would be thankful if you can leave your counterexamples and possibly point out why my proof does not stand.

1) Take an arbitrary $g \in G$. Let $x \in gNg^{-1}$. Then $x = ghg^{-1}$ for some $h \in N$. Then we check whether $x \in N$: $xHx^{-1} = ghg^{-1}Hgh^{-1}g^{-1}$. Can I then conclude that $xHx^{-1} =H$, thus $x \in N$?

2) Assume $H$ is the only $p$-Sylow subgroup of $G$ with order $p^n$ but $H$ is not normal. Then there exists $g \in G$ such that $gHg^{-1} \neq H$. Note that $gHg^{-1}$ is of order $p^n$, contradicting to the assumption that $H$ is the only $p$-Sylow subgroup. What about the other direction? How can I prove it?

Thanks in advance. It would be great if you can write down the details.

1st assertion is false : Consider $S_3$ and any subgroup $H$ of order $2$ in it. Then you can show that $N_{S_3}(H)$ is not normal in $S_3$.

For your second assertion :- you have already done one side. For the reverse side, suppose that $H$ is a normal Sylow-$p$ subgroup of $G$ $\Rightarrow$ $N_G(H)=G$ and hence $|G/N_G(H)|=1$. On the other hand, number of Sylow-$p$ subgroups of $G$ is of the form $1+kp$ and it should also divide $|G/N_G(H)|$. Thus there is a unique Sylow-$p$ subgroup.

• Thanks for replying. I am not familiar with normalizer. May I ask what $|G/N_G(H)|$ is? – Nighty Nov 7 '14 at 15:30
• @LeeKM : it is the cardinality of the set $G/N_G(H)$. If a subgroup is normal, then $N_G(H)=G$. – wanderer Nov 7 '14 at 15:31
• Why do we need to consider $G/N_G(H)$ but not $G$ itself? – Nighty Nov 7 '14 at 15:33
• We don't need to consider!!! But we have some information about the number of Sylow-$p$ subgroups of $G$ from Sylow's 3rd theorem and we also know that it divides $|G/N_G(H)|$. Moreover $H$ is normal, so we know precisely what $N_G(H)$ is. Hence it's quite natural to consider $G/N_G(H)$. – wanderer Nov 7 '14 at 15:36

All $p$-sylow sugroups are conjugate, this is the second Sylow theorem.

• I know but not familiar with the theorem. I just don't see the connection between this and my questions. – Nighty Nov 7 '14 at 14:06

For (1), let $G$ be a simple group and let $H$ be a nontrivial proper subgroup of $G$. What can you say about the normalizer of $H$?

• It is a subgroup of $G$, but it is not normal? – Nighty Nov 7 '14 at 15:22

A little help with Rene Schipperus' hint (can't comment just yet, sorry):

Consider the connection between being a normal subgroup and the conjugates of a subgroup. By definition, if $H$ is a normal Sylow $p$-subgroup of $G$, then $g^{-1}Hg = H$ for every $g \in G$. The conjugates of the subgroup $H$ are of the form $g^{-1}Hg$ for every $g \in G$. What can you conclude from this?