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Prove that $N_G(H)=\{g \in G| gHg^{-1}=H\}$ is the largest subgroup of $G$ such that $H \unlhd N_G(H)$.

I have an idea of the proof that, if we assume $S \leq G$ with $H \unlhd S$ then $$\forall s \in S, \ sHs^{-1}=H$$

We know that for $S$ to be the largest subgroup of $G$ with this property, it should contain every element $g \in G$ with $gHg^{-1}=H$, hence $N_G(H)$ is the largest subgroup of $G$ with this property.

But how could I write this in rigorous mathematical language? Is this a sign that I do not have enough mathematical maturity?

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    $\begingroup$ You might start by defining what you mean by "largest". If you do that properly then the proof is easy. If you do not then you have no hope of writing down a correct proof. $\endgroup$ – Derek Holt Dec 2 '15 at 13:44
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    $\begingroup$ That is not what "largest" means here. It would not even make sense if the groups were infinite. $\endgroup$ – Derek Holt Dec 2 '15 at 13:47
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    $\begingroup$ Try to show (1) that $N_G(H)$ is a subgroup containing $H$ and in which $H$ is normal. (2) any subgroup of $G$ containing $H$ and in which $H$ is normal, must be a subgroup of $N_G(H)$. $\endgroup$ – Nicky Hekster Dec 2 '15 at 13:48
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    $\begingroup$ I do not think that you have grasped the right definition of largest - "proper", "strictly" do not apply here. $\endgroup$ – Nicky Hekster Dec 2 '15 at 13:55
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    $\begingroup$ Largest means that if $K \le G$ and $H \unlhd K$, then $K \le N_G(H)$. $\endgroup$ – Derek Holt Dec 2 '15 at 13:57
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To write this rigorously: Let $K < G$, $H \triangleleft K$. Let $k \in K$. Then, $k Hk^{-1} = H$. Hence, $k \in N_G(H)$ and $K < N_G(H)$.

To say you don't have "enough mathematical maturity" is vague. I don't think anyone can have too much mathematical maturity. I believe your question mainly reflects that you're not comfortable enough with groups. If you study a subject and learn it to the best of your abilities the maturity will follow.

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