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I'm trying to show that $N_G(N_G(H)) = N_G(H)$ for some subgroup H < G. I also understand that the normalizer is described as the largest group which H is normal to, but I don't know how to use that information in a proof.

My attempt

I have shown that $N_G(H) \subseteq N_G(N_G(H))$ since $N_G(H)\lhd N_G(N_G(H)).$

I now want to show that $N_G(N_G(H)) \subseteq N_G(H)$, but I'm getting stuck. Here's what I have so far:

We need to show that if $$x \in N_G(N_G(H)) \Rightarrow x \in N_G(H).$$

i.e. $$xN_G(H)x^{-1} = N_G(H) \Rightarrow xhx^{-1} \in H$$

To show this, I considered taking an element $y \in N_G(H)$, but after that I'm lost. Can anyone provide some insight for how I should continue?

BSplitter
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    This is not true. – Jonathan Hebert Mar 15 '17 at 03:20
  • Can you provide a counterexample? How do you know it's not true? – BSplitter Mar 15 '17 at 03:23
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    If you additionally assume that $H$ is a Sylow $p$-subgroup, then this is true. See http://math.stackexchange.com/questions/67008/normalizer-of-the-normalizer-of-the-sylow-p-subgroup – Kenny Wong Mar 15 '17 at 03:35
  • That's bizarre. The problem statement that I was assigned was, "Prove that for any subgroup H, $N_G(N_G(H)) = N_G(H)$." – BSplitter Mar 15 '17 at 03:40
  • In the interest of providing you with a counter-example, I think if we take $G = D_{3} = {I, F_{1}, F_{2}, F_{3}, R^{120^{\circ}}, R^{240^{\circ}} }$, and $H = {I,F_{1} }$, I believe we have $N_{G}(H) = {I,F_{1}, R^{120^{\circ}}, R^{240^{\circ}} }$, and $N_{G}(N_{G}(H)) = G$, but don't quote me on that. – Jonathan Hebert Mar 15 '17 at 03:49
  • I'm slightly unfamiliar with the notation here. Would $F_2$ be the same as $R^{120^o}F_1?$ – BSplitter Mar 15 '17 at 03:58
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    @JonathanHebert Your example does not work. A group of order $6$ cannot have a subgroup of order $4$. A similar example with $D_4$ works. – Derek Holt Mar 15 '17 at 04:08
  • Woops, sorry! That's what I get for going through that too quickly. – Jonathan Hebert Mar 15 '17 at 04:20
  • I don't believe that the example with $D_4$ works either. The normalizer of the subgroup ${I, F_1}$ is the subgroup itself, so the normalizer of the normalizer will be the same as the normalizer. – BSplitter Mar 15 '17 at 04:28
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    That's not true, the normalizer in $D_4$ of ${1,F_1}$ has order $4$ and includes the rotation through $180^\circ$. – Derek Holt Mar 15 '17 at 05:28
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    Oh yes! I believe that is correct. Working this out, with $H = {1, F}$ we can conclude that $N_G(H) = {1, F, R^{180^o}, R^{180^o}F}$ and $N_G(N_G(H)) = D_8$. Thank you! – BSplitter Mar 15 '17 at 05:43

1 Answers1

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For the sake of completion, I figured that I would go ahead and place an answer to my question here.

The proposition stated in the question is false, unless (as Kenny Wong stated) H is a Sylow p-subgroup. The link to a proof of that is located here: Normalizer of the normalizer of the sylow $p$-subgroup.

To show that the the statement is false, examine the counterexample $D_8$, i.e. the set of symmetries on a square. (Sometimes, this is denoted as $D_4$).

$$G := D_8 = \{e,\ r,\ r^2,\ r^3,\ f,\ fr,\ fr^2,\ fr^3\}$$ where $r$ is a 90 degree rotation, and f is a flip. Let $H$ be $\{e,\ f\}$. Obviously, $H < G$.

$$N_G(H) = \{e,\ f,\ r^2,\ fr^2\}$$ $$\text{Then, }N_G(N_G(H)) = D_8.$$ $$\text{So, }N_G(H) \neq N_G(N_G(H)).$$

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