Centralizer and Normalizer as Group Action

Question

Could someone explain the following to me? I understand all the terms used in the text but have no intuition of what it's saying.

Dummit and Foote Pg. 52

Answer 1

A (left) group action of a group $G$ on a set $X$ is a map $G\times X\to X$ satisfying:

• $x^e = x$ for any $x\in X$
• $(x^g)^h = x^{hg}$

where the image of $(g,x)$ is denoted as $x^g$.

There are two important subgroups of $G$ that arise from a group action: the kernel and stabilizers. The kernel of the action is the set of all $g\in G$ such that $x^g = x$ for all $x\in X$. Given a subset $S\subset X$, the stabilizer of $S$ denoted by $G_S$ is the set of all $g\in G$ such that $s^g\in S$ for all $s\in S$.

Note that the kernel collects group elements that stabilize every individual element in $X$, and the stabilizer collects group elements that stabilize a particular set of elements in $X$. You should check for yourself that these are subgroups!

Let’s focus on the particular example where you give me any group $G$, we take $X$ to be $\mathcal{P}(G)$, and given any $S\subseteq G$ and $g\in G$ define $S^g = gSg^{-1} = \{gsg^{-1}: s\in S\}$. The stabilizer subgroup we defined above for this action on some set $A\subseteq G$ is the set of all $g\in G$ such that $gAg^{-1} = A$ — which is exactly the normalizer subgroup $N_G(A)$! Thus we know that the normalizer is a subgroup because stabilizers are.

Now, take our group to be $N_G(A)$ and take $X$ to be $A$, with the same action as above. The kernel of this action is the set of all $g\in N_G(A)$ such that $a^g = gag^{-1} = a$ for all $a\in A$, i.e. the elements in $N_G(A)$ that commute with all elements in $A$. This is exactly the centralizer $C_G(A)$ of $A$ in $G$, and so it must be a subgroup because kernels are!

Thus $C_G(A)$ is a subgroup of $N_G(A)$ since it is the kernel of our second action, and $N_G(A)$ is a subgroup of $G$ because it is the stabilizer of $A$ in our first action. Thus $C_G(A) \le N_G(A) \le G$.

This is said in your text, but explain to yourself why $Z(G)$ is a subgroup of $G$ by think of $Z(G)$ as the kernel of a group action!

TL;DR If you understand kernels and stabilizers of group actions, many important objects pop out as having this form.

Answer 2

It's saying that if you want to prove that centralizers, normalizers, and kernels are subgroups of G, it is enough to show that they are stabilizers or kernels of group actions. Then they exhibit a particular group action (conjugation) and show that $N_G(A)$ is the stabilizer of A and Z(G) is the kernel of this action. Thus those two sets are subgroups. Then they let $N_G(A)$ act only on A and show the $C_G(A)$ is the kernel of this second action and thus a subgroup.