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, C_G(A) = {g ∈ G | gag^-1 = a ∀a ∈ A} = {g ∈ G | ga = ag ∀ a ∈ A}.

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

Proof.

1. C_G(A) ≠ ∅ because e ∈ C_G(A), ea = ae ∀a ∈ A.
2. ∀y ∈ C_G(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 ∈ C_G(A)
3. ∀x, y ∈ C_G(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 ∈ C_G(A)] xax^-1 = [x ∈ C_G(A)] a ⇒ xy ∈ C_G(A).

Example. In an Abelian group G, C_G(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) = C_G(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 N_G(A) = {g ∈ G: gAg^-1 = A} = {g ∈ G: gA = Ag}. Clearly C_G(A) ⊆ N_G(A) and both are subgroups of G.

Examples.

• The center of an Abelian group, G, is the group itself. C_G(A) = N_G(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 = aⁱb^j for some i ∈ {0, 1, 2, 3} and j ∈ {0, 1}, x ∈ Z(Q₈) ↭ xa = ax ∧ xb = bx

xa = ax ↭ aⁱb^ja = aaⁱb^j ↭ aⁱb^ja = aⁱ⁺¹b^j ↭ [By cancellation laws] b^ja = ab^j.

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: ba^k = 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 D_n = {α, β: αⁿ = β² = e, βαβ = α^-1}, n ≥ 3.

x ∈ ℤ(D_n) iff xα = αx and xβ = βx.

x ∈ ℤ(D_n) = ⟨α, β⟩ ⇒ x = αⁱβ^j ⇒ xα = αx ↭ αⁱβ^jα = ααⁱβ^j = αⁱ⁺¹β^j ↔ αⁱβ^jα = αⁱ⁺¹β^j ↭ [Cancellation laws] β^jα = αβ^j

In particular j = 1, βα = αβ [βαβ = α^-1 ⇒ βα = 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(D_n) ⇒ xβ = βx ⇒ αⁱβ = βαⁱ ⇒ [Property: α^kβ = βα^n-k ↭ α^n-kβ = βα^k] αⁱβ = α^n-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(D_n) = $ \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 C_D8(A) = A. We know that all powers of r commute with each other ⇒ A ⊆ C_D8(A). s ∉ C_D8(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 ∈ C_D8(A) because C_D8(A) ≤ G, then s = (srⁱ)(r^-i) ∈ C_D8(A) -closure- ⊥ Therefore, C_D8(A) = A

Let’s prove that N_D8(A) = D₈. First, C_D8(A) = A ⊆ N_D8(A).

sAs^-1 = {s1s^-1, srs^-1, sr²s^-1, sr³s^-1} =_🔑 {1, r³, r², r} = A ⇒ s ∈ N_D8(A).

🔑 srs^-1 = [sr = r^-1s] r^-1ss^-1 = r^-1 = [r⁴=1] r³. sr²s^-1 = sr²ss^-2 [r^ks = sr^4-k] ssr²s^-2 = [s² = s^-2 = e] r². sr³s^-1 = sr³ss^-2 = s²rs^-2 = r.

r, s ∈ N_D8(A) ≤ D₈ ⇒ [Every subgroup is closed under the group operation] sⁱr^j ∈ N_D8(A) ⇒ N_D8(A) = D₈.

Bibliography

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.