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.
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.
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, Q_{8} is {e, a^{2}}.
Q_{8} = ⟨a, b: a^{4} = e, b^{2} = a^{2}, aba = b⟩, so in particular Q_{8} is generated by both a and b.
Let x ∈ Q_{8}, x = a^{i}b^{j} for some i ∈ {0, 1, 2, 3} and j ∈ {0, 1}, x ∈ Z(Q_{8}) ↭ xa = ax ∧ xb = bx
xa = ax ↭ a^{i}b^{j}a = aa^{i}b^{j} ↭ a^{i}b^{j}a = a^{i+1}b^{j} ↭ [By cancellation laws] b^{j}a = ab^{j}.
For j = 1, ba = ab ⇒ [aba = b ⇒ ab = ba^{-1}] ba = ba^{-1} ⇒ a = a^{-1} ⇒ a^{2} = e⊥ (the order of a is 4, not 2) ⇒ x = a^{i} for some i ∈ {0, 1, 2, 3}.
xb = bx ↭ a^{i}b = ba^{i} ↭ [Product of Generating Elements of Quaternion Group, Q_{8} satisfies: ba^{k} = a^{-k}b.] a^{i}b = a^{-i}b ↭ a^{i} = a^{-i} ↭ a^{2i} = e = a^{4}. Therefore, either i = 0 (x = a^{0} = e) or 2i = 4, x = a^{i} = a^{2}.
x ∈ ℤ(D_{n}) iff xα = αx and xβ = βx.
x ∈ ℤ(D_{n}) = ⟨α, β⟩ ⇒ x = α^{i}β^{j} ⇒ xα = αx ↭ α^{i}β^{j}α = αα^{i}β^{j} = α^{i+1}β^{j} ↔ α^{i}β^{j}α = α^{i+1}β^{j} ↭ [Cancellation laws] β^{j}α = αβ^{j}
In particular j = 1, βα = α^{}β [βαβ = α^{-1} ⇒ βα = α^{-1}β^{-1} = αβ] Taking into consideration only the last equality, α^{-1}β^{-1} = αβ ⇒ [β^{2} = e ⇒ β = β^{-1}] α^{-1}β = αβ ⇒ [Cancellation laws] α^{-1} = α ⇒ α^{2} = e ⊥. Therefore, x cannot be a reflection (all reflections are generated by α^{i}β, 0 ≤ i ≤ n-1, j = 1), it necessary needs to be a rotation, x = α^{i}.
x ∈ ℤ(D_{n}) ⇒ xβ = βx ⇒ α^{i}β = βα^{i} ⇒ [Property: α^{k}β = βα^{n-k} ↭ α^{n-k}β = βα^{k}] α^{i}β = α^{n-i}β = [α^{n} = e] α^{-i}β ⇒ α^{i}β = α^{-i}β ⇒ [Cancellation laws] α^{i} = α^{-i} ⇒ α^{2i} = e ⇒ i = 0 (x = α^{0} = 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}$
Let’s prove that N_{D8}(A) = D_{8}. First, C_{D8}(A) = A ⊆ N_{D8}(A).
sAs^{-1} = {s1s^{-1}, srs^{-1}, sr^{2}s^{-1}, sr^{3}s^{-1}} =_{🔑} {1, r^{3}, r^{2}, r} = A ⇒ s ∈ N_{D8}(A).
🔑 srs^{-1} = [sr = r^{-1}s] r^{-1}ss^{-1} = r^{-1} = [r^{4}=1] r^{3}. sr^{2}s^{-1} = sr^{2}ss^{-2} [r^{k}s = sr^{4-k}] ssr^{2}s^{-2} = [s^{2} = s^{-2} = e] r^{2}. sr^{3}s^{-1} = sr^{3}ss^{-2} = s^{2}rs^{-2} = r.
r, s ∈ N_{D8}(A) ≤ D_{8} ⇒ [Every subgroup is closed under the group operation] s^{i}r^{j} ∈ N_{D8}(A) ⇒ N_{D8}(A) = D_{8}.