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Let $H$ be a subgroup of a group $G$ and let $N(H)$ be its normalizer

prove that

H is a normal subgroup of $N(H)$

def of normalizeer $$N_G(H)= \{ g \in G : gH=Hg \} $$

attempt 1 (its wrong)

Need to show three things first two that it is a subgroup that it is that it closed and it has the inverse thirdly that it is normal

1) trying to show that it for any elment it has its inverese

$g\in N_g(H)$ is $g^{-1}\in N(H)$?

so $g \in N_g(H)$ that is $gH=Hg \iff gHg^{-1}=H$

somehow $$\begin{aligned} steps??? \\ \vdots \\ g^{-1}H =Hg^{-1} \end{aligned} $$

2) that it is closed if $g_1,g_2 \in N(H)$ is $g_1 g_2 \in N(H)$

that is if

$$ \begin{aligned} g_1 \in N_G(H) \Rightarrow g_1H =H g_1 g_2 \in N_G(H) \Rightarrow g_2H =H g_2 \end{aligned} $$

need to show that $g_1 g_2 \in N_g(H)$

3) need to show that for any $g \in G$ $g^{-1}N(H)g=N(H)$ $g^{-1}N(H)g = N(H)g^{-1}g=N(H) $

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    You seem to be trying to prove $N(H)$ is a normal subgroup of $G$, not that $H$ is a normal subgroup of $N(H)$. – 2017-02-28
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    $h^{-1} H h = h^{-1} h H =H$ is that all ?? – 2017-02-28
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    do I need to show that H is a subgroup of N(H) when its already a subgroup of $G$?? – 2017-02-28

1 Answers 1

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Well, since for any $\;h\in H\;$ we trivially have $\;hH=H=Hh\;$ , then we get that $\;h\in N_G(H)\implies H\subset N_G(H)\;$ .

Now,

$$H\lhd N_G(H)\iff \,\forall\,n\in N_G(H)\;,\;\;H^n=H\iff Hn=nH$$

but this is exactly the definition of $\;N_G(H)\;$ ...! Now we can deduce that $\;N_G(H)\;$ is the maximal subgroup of $\;G\;$ containing $\;H\;$ in which $\;H\;$ is normal, and thus $\;H\lhd G\iff N_G(H)=G\;$ .