# Centralizer : center subgroup ~ normalizer : normal subgroup ~ ?: quotient group

Centralizer subgroup: $$C_G(A)=\{g\in G\mid gag^{-1}=a,\forall a\in A\}$$ Center subgroup: $$Z(G)=\{g\in G\mid ga=ag,\forall a\in G\}$$

Centralizer and center are subgroups of $$G$$. The centralizer takes a subset $$A\in G$$ and the center always uses the entire group $$G$$. So the centralizer becomes center when $$A=G$$ $$C_G(G)= Z(G).$$

My questions are that:

1. Under what precise mathematical conditions that the normalizer $$N_G(A)$$ equal to a normal subgroup $$N$$ of $$G$$? When is $$N_G(A)$$ not equal to a normal subgroup $$N$$? See for instances.

2. Since the quotient group is the $$G/N$$, are there analogous concepts of quotient (not a subgroup) for $$G/N_G(A)$$, important for what properties or theorems?

In short, I am exhausting the concepts of

Centralizer : center subgroup ~ normalizer : normal subgroup ~ ?: quotient group

The normalizer works "the other way around" to what you may think it does. The normalizer of $$A$$ in $$G$$ is the largest subgroup of $$G$$ that contains $$A$$ and in which $$A$$ is normal.
That is, $$N_G(A) =\{g\in G:gA=Ag\}$$. By its very definition, $$A$$ is normal in $$N_G(A)$$. However, $$N_G(A)$$ may not be a normal subgroup of $$G$$, it seems you already have examples of this. I am not sure you should expect any "reasonably general" condition that guarantees a normalizer is normal, but rather, the other way around.
One way in which normalizers and centralizers interact with quotients is as follows: each element $$g\in N_G(A)$$ defines an automorphism of $$A$$ by conjugation, and hence there is a map $$N_G(A) \longrightarrow \mathrm{Aut}(A)$$ that assigns $$g$$ to the automorphism $$a\longmapsto {}^ga$$. By definition, the kernel of this map is the centralizer of $$A$$ in $$G$$, so one obtains the "NC lemma" that $$C_G(A)$$ is normal in $$N_G(A)$$ and that $$N_G(A)/C_G(A)$$ is isomorphic to a subgroup of $$\mathrm{Aut}(A)$$.