0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for replying. I am not familiar with normalizer. May I ask what $|G/N_G(H)|$ is? $\endgroup$ – Nighty Nov 7 '14 at 15:30
  • $\begingroup$ @LeeKM : it is the cardinality of the set $G/N_G(H)$. If a subgroup is normal, then $N_G(H)=G$. $\endgroup$ – wanderer Nov 7 '14 at 15:31
  • $\begingroup$ Why do we need to consider $G/N_G(H)$ but not $G$ itself? $\endgroup$ – Nighty Nov 7 '14 at 15:33
  • $\begingroup$ 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 $G/N_G(H)$. $\endgroup$ – wanderer Nov 7 '14 at 15:36
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I know but not familiar with the theorem. I just don't see the connection between this and my questions. $\endgroup$ – Nighty Nov 7 '14 at 14:06
0
$\begingroup$

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 $H$?

$\endgroup$
  • $\begingroup$ It is a subgroup of $G$, but it is not normal? $\endgroup$ – Nighty Nov 7 '14 at 15:22
0
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.