Centralizers and Normalizers

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On this page:

• Introduction
• Centralizers and Normalizers 101
• Basic Examples of $C_G(A)$, $N_G(A)$, and $Z(G)$
• Worked Exercises

Introduction

Now that we have the language of group actions, orbits, and stabilizers, we will slowly begin to turn our attention to specific group actions (the conjugation action) and study their stabilizers—soon to be called centrailizers and normalizers.

Centralizers and Normalizers 101

Commutativity is “good” because we like it when $xy = yx$ — the structure is way easier to understand and pathological or complicated examples are less likely to arise. But, in general, we know that many groups do not like to play nice in this way so we have to refine our commutativity assumptions a bit.

[definition] [deftitle]Definition (Centralizer).[/deftitle] Let $G$ be a group and $A$ be a subset of $G$. We call the subset

of $G$ the centralizer of $A$ in $G$. In the case where $A = \{a\}$, we define $C_G(a) \coloneqq C_G(\{a\})$. [/definition]

[theorem] [thmtitle]Proposition ($C_G(A) \le G$).[/thmtitle]

Let $G$ be a group and $A \subseteq G$. Then $C_G(A) \le G$. [/theorem]

[proof] It is immediate that $1 \in C_G(A)$. Now, if $g, h \in C_G(A)$, then

for all $a\in A$, where the last equality follows from

Thus, $h^{-1}g \in C_G(A)$. [/proof]

Another perspective that will ultimately be more useful is that the relation $ga = ag$ is equivalent to $gag^{-1} = a$. So we can reexpress the centralizer as

So, another way we could have proved that $C_G(A)$ is a subgroup is to realize that it is the stabilizer of a conjugation action under an appropriate context. Notice that $G$-conjugation on $A$ is not necessarily an action since $gag^{-1}$ might leave $A$, so we need to impose the addition restriction that $gag^{-1} \in A$ before thinking of $C_G(A)$ as the kernel of a conjugation action. This is where the normalizer comes in.

[definition] [deftitle]Definition (Normalizer).[/deftitle]

Let $G$ be a group and $A \subseteq G$. The set

is called the normalizer of $A$ in $G$. [/definition]

The proof that $N_G(A)$ is a subgroup of $G$ is pretty much the same as the proof that $C_G(A)$ is a subgroup of $G$ so we omit the proof of the following proposition.

[theorem] [thmtitle]Proposition ($N_G(A) \le G$).[/thmtitle]

Let $G$ be a group and $A \subseteq G$. Then $N_G(A) \le G$. [/theorem]

Another straightforward result is that $C_G(A) \le N_G(A)$.

[theorem] [thmtitle]Proposition ($C_G(A) \le N_G(A) \le G$).[/thmtitle]

Let $G$ be a group and $A \subseteq G$. Then

[/theorem]

The next theorem is another way of restating the definition and so we omit its proof.

[theorem] [thmtitle]Theorem ($N_G(A) = \Stab(A)$ under action of $G$ on $2^G$).[/thmtitle]

Let $G$ be a group and $2^G$ the power set of $G$. Furthermore, let $G$ act on $2^G$ by conjugation:

Then the stabilizer of $A$ under this $G$-action is precisely $N_G(A)$. [/theorem]

And, finally, now we express $C_G(A)$ in terms of group actions. Again, the result is really just another way of stating the definition phrased in terms of group actions, so we omit the proof.

[theorem] [thmtitle]Theorem ($C_G(A)$ as kernel of $N_G(A)$-conjugation of $G$).[/thmtitle]

Let $G$ be a group and $A \subseteq G$. Let $N_G(A)$ act on the set $A$ by conjugation. Then $C_G(A)$ is the kernel of this action. [/theorem]

We will finish off this section with one final definition that is important in group theory.

[definition] [deftitle]Definition (Center of group).[/deftitle]

Let $G$ be a group. Then the center $Z(G)$ of $G$ is the kernel of $G$ acting on itself by conjugation. That is,

[/definition]

[example] [extitle]Remark.[/extitle]

Another way we can also think of $Z(G)$ is that $Z(G) = C_G(G)$. [/example]

Basic Examples of $C_G(A)$, $N_G(A)$, and $Z(G)$

[example] [extitle]Abelian groups.[/extitle]

Let $G$ be an abelian group. Then all elements of $G$ commute so

for any $A \subseteq G$. [/example]

[example] [extitle]Dihedral groups.[/extitle]

Consider the Dihedral group

Then we claim that

This is easy to see by doing some casework. Of course, $r^{n/2}$ commutes with powers of $r$ and $r^{n/2} = r^{-n/2}$ is implied by the relation $r^n = 1$. So

and a very similar argument shows that

Similar casework is used to show that $Z(D_{2n})$ cannot contain any other elements of $D_{2n}$.

A classical example of centralizers and normalizers is to let

Then more case by case analysis shows that $C_{D_{2n}}(A) = A$. More interesting is

This can be easily seen by noting that the relation $srs = r^{-1}$ implies that conjugation involving terms with $s$ just sends you back to $A$ (this can be easily seen by experimentation). [/example]

Worked Exercises

[example] [extitle]Dummit and Foote Problem 2.2.2.[/extitle]

Let $G$ be a group. Then

[/example]

[proof] ($C_G(Z(G)) = G$) The center of $G$ commutes with everything in $G$ so the claim follows immediately.

($N_G(Z(G)) = G$) Same argument. [/proof]

[example] [extitle]Dummit and Foote Problem 2.2.3[/extitle]

If $A$ and $B$ are subsets of the group $G$ with $A \subseteq B$, then $C_G(B)$ is a subgroup of $C_G(A)$. [/example]

[proof] Since $A$ is contained inside $B$, anything $g \in G$ that commutes with all elements of $B$ must also commute with the elements of $A$. So the claim follows. [/proof]

[example] [extitle]Dummit and Foote Problem 2.2.10[/extitle]

Let $H$ be a subgroup of order $2$ in $G$. Then $N_G(H) = C_G(H)$. In particular, if $N_G(H) = G$ then $H \le Z(G)$. [/example]

[proof] Since we already know that $C_G(H) \le N_G(H)$, it suffices to show that $N_G(H) \subseteq C_G(H)$. Let $h$ be the element of $H$ that has order $2$. Then, if $g\in N_G$, we know that $ghg^{-1} = h$. If that were not the case, then $gHg^{-1}$ would be smaller than $H$. So $g \in C_G(H)$ as desired. To prove the remaining claim, we note that since $N_G(H) = C_G(H)$, it follows that every $g \in G$ commutes with the elements of $H$ — which also means that every element of $H$ commutes with every element of $G$. Thus, $H \subseteq Z(G)$ and the claim follows. [/proof]