    # Center, Centralizers, & Normalizers.

Beneath this mask there is more than flesh. Beneath this mask there is an idea, Mr. Creedy, and ideas are bulletproof, V for Vendetta Definition. Let G be a group, and A let be a non-empty subset of G.The centralizer of a group G is the subset of elements of G that commute with every element of A. Formally, CG(A) = {g ∈ G | gag-1 = a ∀a ∈ A} = {g ∈ G | ga = ag ∀ a ∈ A}.

Proposition. CG(A) is a subgroup of G.

Proof.

1. CG(A) ≠ ∅ because e ∈ CG(A), ea = ae ∀a ∈ A.
2. ∀y ∈ CG(A) ⇒ yay-1=a, ∀a ∈ A ⇒ [By multiplying both sides by y-1] y-1(yay-1) = y-1a ⇒ [Associativity] ay-1 = y-1a ⇒ [By multiplying both sides by y] ay-1y = y-1ay ⇒ a = y-1ay ∀a ∈ A ⇒ y-1 ∈ CG(A)
3. ∀x, y ∈ CG(A), ∀a ∈ A, (xy)a(xy)-1 = [shoes and socks principle] (xy)a(y-1x-1) = [By the associative law] x(yay-1)x-1 = [y ∈ CG(A)] xax-1 = [x ∈ CG(A)] a ⇒ xy ∈ CG(A).

Example. In an Abelian group G, CG(A) = G, for all subsets A.

Definition. The center of a group, G, is the set of elements that commute with all the elements of the group. Z(G) = {g ∈ G | gx = xg, ∀x ∈ G}.

Proposition. Z(G) is a subgroup of G, Z(G) ≤ G.

Proof: Z(G) = CG(G), so the argument previously shown proves Z(G) ≤ G.

As an exercise, let’s prove it directly with the Two-Step Subgroup Test.

1. ∀ a, b ∈ Z(G), ab ∈ Z(G) ↭ (ab)x = x(ba), ∀x ∈ G. (ab)x = [a ∈ Z(G)] (ba)x = [Associativity] b(ax) = [a ∈ Z(G)] b(xa) = [Associativity] (bx)a = [b ∈ Z(G)] (xb)a = [Associativity] x(ba)
2. ∀ a ∈ Z(G), a-1∈ Z(G)? ∀x ∈ G, a-1x = xa-1? ↭[Cancellation law property] (a-1x)a = (xa-1)a ↭[By associative and identity properties] a-1(xa) = x ↭ a(a-1xa) = ax ↭[By associative and identity properties] xa = ax, and that is true because a ∈ Z(G)∎

gAg-1 = {gag-1 | a ∈ A}. The normalizer of A in the group G is defined as NG(A) = {g ∈ G: gAg-1 = A} = {g ∈ G: gA = Ag}. Clearly CG(A) ⊆ NG(A) and both are subgroups of G.

Examples.

• The center of an Abelian group, G, is the group itself. CG(A) = NG(A) = G for any subset A of G.

• The center of the quaternion group, Q8 is {e, a2}.

Q8 = ⟨a, b: a4 = e, b2 = a2, aba = b⟩, so in particular Q8 is generated by both a and b.

Let x ∈ Q8, x = aibj for some i ∈ {0, 1, 2, 3} and j ∈ {0, 1}, x ∈ Z(Q8) ↭ xa = ax ∧ xb = bx

xa = ax ↭ aibja = aaibj ↭ aibja = ai+1bj ↭ [By cancellation laws] bja = abj.

For j = 1, ba = ab ⇒ [aba = b ⇒ ab = ba-1] ba = ba-1 ⇒ a = a-1 ⇒ a2 = e⊥ (the order of a is 4, not 2) ⇒ x = ai for some i ∈ {0, 1, 2, 3}.

xb = bx ↭ aib = bai ↭ [Product of Generating Elements of Quaternion Group, Q8 satisfies: bak = a-kb.] aib = a-ib ↭ ai = a-i ↭ a2i = e = a4. Therefore, either i = 0 (x = a0 = e) or 2i = 4, x = ai = a2.

• Let’s calculate the center of the dihedral group Dn = {α, β: αn = β2 = e, βαβ = α-1}, n ≥ 3.

x ∈ ℤ(Dn) iff xα = αx and xβ = βx.

x ∈ ℤ(Dn) = ⟨α, β⟩ ⇒ x = αiβj ⇒ xα = αx ↭ αiβjα = ααiβj = αi+1βj ↔ αiβjα = αi+1βj ↭ [Cancellation laws] βjα = αβj

In particular j = 1, βα = αβ [βαβ = α-1βα = α-1β-1 = αβ] Taking into consideration only the last equality, α-1β-1 = αβ ⇒ [β2 = e ⇒ β = β-1] α-1β = αβ ⇒ [Cancellation laws] α-1 = α ⇒ α2 = e ⊥. Therefore, x cannot be a reflection (all reflections are generated by αiβ, 0 ≤ i ≤ n-1, j = 1), it necessary needs to be a rotation, x = αi.

x ∈ ℤ(Dn) ⇒ xβ = βx ⇒ αiβ = βαi ⇒ [Property: αkβ = βαn-k ↭ αn-kβ = βαk] αiβ = αn-iβ = [αn = e] α-iβ ⇒ αiβ = α-iβ ⇒ [Cancellation laws] αi = α-i ⇒ α2i = e ⇒ i = 0 (x = α0 = e) or i = n/2 and n is even.

Therefore, ℤ(Dn) = $\begin{cases} e, n~is~ odd\\\\ e~ and~ α^{\frac{n}{2}}, n~is~ even \end{cases}$

• G = D8, A = {1, r, r2, r3}. Let’s prove that CD8(A) = A. We know that all powers of r commute with each other ⇒ A ⊆ CD8(A). s ∉ CD8(A) because ∃a ∈ A (a=r): sr = r-1s ≠ rs. Finally, all elements of D8 that are not in A are all of the form sri, 0 ≤ i ≤ 3. If there were some i such that sri ∈ CD8(A) because CD8(A) ≤ G, then s = (sri)(r-i) ∈ CD8(A) -closure- ⊥ Therefore, CD8(A) = A

Let’s prove that ND8(A) = D8. First, CD8(A) = A ⊆ ND8(A).

sAs-1 = {s1s-1, srs-1, sr2s-1, sr3s-1} =🔑 {1, r3, r2, r} = A ⇒ s ∈ ND8(A).

🔑 srs-1 = [sr = r-1s] r-1ss-1 = r-1 = [r4=1] r3. sr2s-1 = sr2ss-2 [rks = sr4-k] ssr2s-2 = [s2 = s-2 = e] r2. sr3s-1 = sr3ss-2 = s2rs-2 = r.

r, s ∈ ND8(A) ≤ D8 ⇒ [Every subgroup is closed under the group operation] sirj ∈ ND8(A) ⇒ ND8(A) = D8.

# Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.
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