# How to write a rigorous proof for normalisers $N_{G}(H)$ being the largest subgroups of $G$ such that $H \unlhd N_{G}(H)$

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?

• 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. – Derek Holt Dec 2 '15 at 13:44
• That is not what "largest" means here. It would not even make sense if the groups were infinite. – Derek Holt Dec 2 '15 at 13:47
• 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 $N_G(H)$. – Nicky Hekster Dec 2 '15 at 13:48
• I do not think that you have grasped the right definition of largest - "proper", "strictly" do not apply here. – Nicky Hekster Dec 2 '15 at 13:55
• Largest means that if $K \le G$ and $H \unlhd K$, then $K \le N_G(H)$. – Derek Holt Dec 2 '15 at 13:57

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 $K < N_G(H)$.