Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup

Theorem
Let $G$ be a group.

Let $H$ be a subgroup of $G$.

Then $N_G \left({H}\right)$, the normalizer of $H$ in $G$, is the largest subgroup of $G$ containing $H$ as a normal subgroup.

Proof
From Subgroup is Subgroup of Normalizer, we have that $H \le N_G \left({H}\right)$.

Now we need to show that $H \triangleleft N_G \left({H}\right)$.

For $a \in N_G \left({H}\right)$, the conjugate of $H$ by $a$ in $N_G \left({H}\right)$ is:

so:
 * $\forall a \in N_G \left({H}\right): H^a = H$

and so by definition of normal subgroup, $H \triangleleft N_G \left({H}\right)$.

Now we need to show that $N_G \left({H}\right)$ is the largest subgroup of $G$ containing $H$ such that $H \triangleleft N_G \left({H}\right)$.

That is, to show that any subgroup of $G$ in which $H$ is normal is also a subset of $N_G \left({H}\right)$.

Take any $N$ such that $H \triangleleft N \le G$.

In $N$, the conjugate of $H$ by $a \in N$ is $N \cap H^a = H$.

Therefore $H \subseteq H^a$.

Similarly, $H \subseteq H^{a^{-1}}$, so $H^a \subseteq \left({H^a}\right)^{a^{-1}} = H$.

Thus, $\forall a \in N: H^a = H, a \in N_G \left({H}\right)$.

That is, $N \subseteq N_G \left({H}\right)$.

So what we have shown is that any subgroup of $G$ in which $H$ is normal is a subset of $N_G \left({H}\right)$, which is another way of saying that $N_G \left({H}\right)$ is the largest such subgroup.