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) $
Best Answer
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 $\;H\lhd G\iff N_G(H)=G\;$ .