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 $$H \triangleleft N_G \left({H}\right)$$.

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


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

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.