Conjugacy Classes. The Class Equation.

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Definition. Any two elements a and b of a group G are said conjugate if xax^-1 = b for some x ∈ G.

Conjugacy is an equivalent relation on G, and the conjugacy class of “a”, denoted by cl(a) ={xax^-1 | x ∈ G} is the equivalence class of “a” under conjugacy, and therefore we could partition any group G into disjoint conjugacy classes.

Example. In D₄, cl(H) = {R₀HR₀^-1 (= H), R₉₀HR₉₀^-1 (= V), R₁₈₀HR₁₈₀^-1 (= H), R₂₇₀HR₂₇₀^-1 (= V), HHH^-1, VHV^-1, DHD^-1, D’HD’^-1} = {H, V}. Similarly, cl(R_o) = {R₀}, cl(R₉₀) = cl(R₂₇₀) = {R₉₀, R₂₇₀}, etc.

Theorem. Let G be a finite group and let g be any element of G. Then, the number of conjugates of g is the index of the centralizer, that is, |cl(g)| = |G:Z(g)|

Proof.

Recall that H = Z(g), also denoted as C(g), is the set of all the elements that actually commute with this given element “g”, that is, {x ∈ G | xg = gx}.

Let R = G/C(g) {right cosets of C(g) in G}, and define Φ:G/C(g) → cl(g), Φ(C(g)x) = x^-1gx.

• Φ is well-defined. If C(g)x = C(g)y ⇒ xy^-1 ∈ C(g) ⇒ (xy^-1)g = g(xy^-1) ⇒ [*x^-1] y^-1g = x^-1gxy^-1 [*y] = y^-1gy = x^-1gxy^-1y ⇒ y^-1gy = x^-1gx.
• Φ is one-to-one. Suppose Φ(C(g)x) = Φ(C(g)y) ⇒ We need to show that C(g)x = C(g)y, that is, xy^-1∈ C(g) or, alternatively, (xy^-1)g = g(xy^-1)

Φ(C(g)x) = Φ(C(g)y) ⇒ x^-1gx = y^-1gy ⇒ [*y^-1] x^-1gxy^-1 = y^-1gyy^-1 ⇒ x^-1gxy^-1 = y^-1g ⇒ [*x] xx^-1gxy^-1 = xy^-1g ⇒ gxy^-1 = xy^-1g

• Φ is onto. Since both sets are finite sets (G is a finite group), this implies that Φ is bijective.

Therefore, Φ is isomorphism ⇒ |R| = |G:Z(g)| = |cl(g)| ∎

Corollary. If G is a finite group, then |cl(a)| divides |G|.

Proof.

Recall that the centralizer of a in G is a subgroup of G ⇒ the set of left (or right) cosets G:C(a) divides the order of G by Lagrange’s Theorem, and therefore |cl(a)| =[previous theorem, Z(a) is also denoted as C(a)] |G:Z(a)| divides the order of G.

The Class Equation. Let G be a nontrivial finite group. Suppose that a₁, a₂, ···, a_k are the representatives of the conjugacy classes that have size > 1. Then, $|G| = |Z(G)|+\sum_{i=1}^k |G:C(a_k)|$

Proof.

We have already demonstrated that |cl(g)| = |G:C(g)|, and C(g) is a subgroup of G, then |G| = |G:C(a_i)|·|C(a_i)|, then |G|/|C(a_i)| = |cl(a_i)| = |G:C(a_i)|

Since the conjugacy classes partition a group G, let b₁, ··· b_r be representatives of the conjugacy classes that have size 1

G = cl(b₁) ∪ ··· cl(b_r) ∪ cl(a₁) ∪ ··· cl(a_k) ⇒ [A partition is a union of disjoint classes] |G| = |cl(b₁)| + ··· + |cl(b_r)| + |cl(a₁)| + ··· + |cl(a_k)| = [Previous result: a∈ C(G)=Z(G) iff cl(a)={a}, in other words, the center of G is the set of all representatives of the conjugacy classes that have size 1] 1 + ··· 1 + |cl(a₁)| + ··· + |cl(a_k)| = |Z(G)| + |G:C(a₁)| + ··· + |G:C(a_k)| ⇒ $|G| = |Z(G)|+\sum_{i=1}^k |G:C(a_k)|$

Lemma. If G/Z(G) is cyclic, then G is Abelian.

Proof. (Based on Mathonline)

Recall that Z(G) ◁ G ⇒ the quotient G/Z(G) is well-defined ⇒[By assumption, G/Z(G) is cyclic] ∃g ∈ G: G/Z(G) = ⟨gZ(G)⟩

Let h ∈ G ⇒ hZ(G) ∈ G/Z(G) = ⟨gZ(G)⟩ ⇒ ∃n ∈ ℤ: hZ(G) = (gZ(G))ⁿ = gⁿZ(G), that is, hZ(G) = gⁿZ(G) ↭ (gⁿ)^-1h ∈ Z(G) ⇒ ∃i∈ Z(G): i = (gⁿ)^-1h ⇒ h = gⁿi. In other words, ∀h∈ G, ∃n ∈ ℤ & i∈ Z(G): h = gⁿi

∀h₁, h₂ ∈ G, we claim h₁h₂ = h₂h₁.

We already know ∃n₁, n₂ ∈ ℤ & i₁, i₂ ∈ Z(G): h₁ = g^n₁i₁ and h₂ = g^n₂i₂

h₁h₂ = g^n₁i₁g^n₂i₂ =[i₁ ∈ Z(G)] g^n₁g^n₂i₁i₂ = g^n₁+n₂i₁i₂ = g^n₂g^n₁i₁i₂ =[i₂ ∈ Z(G)] g^n₂g^n₁i₂i₁ =[i₂ ∈ Z(G)] g^n₂i₂g^n₁i₁ = h₂h₁ ∎

Theorem. Every p-group has a nontrivial center.

Proof.

Recall that given a prime number p, a p-group is a group in which the order of every element is a power of p. |G| = pⁿ, p prime and some n > 0.

By the class equation, $|G| = |Z(G)|+\sum_{i=1}^k |G:C(a_k)|$, but recall that |G:C(a_i)|=|G|/|C(a_i)| where C(a_i) is a subgroup, its order must divide |G|, |G|/|C(a_i)| = $\frac{p^n}{p^k}$, k < n.

$|G| - \sum_{i=1}^k |G:C(a_k)| = |Z(G)|$ Therefore, each term on the left is divisible by p, it follows that p also divides |Z(G)| ⇒ |Z(G)| ≠ 1 ∎

Theorem. If G is a group of order p², |G|=p², where p is prime, then G is Abelian.

Proof. By Lagrange and previous theorem, we know that (i) Z(G) is a subgroup of G, (ii) |Z(G)| divides |G|, so |Z(G)| =1, p or p², and (iii) Every p-group has nontrivial center, |Z(G)| = p or p²

1. |Z(G)| = p² ⇒ G = Z(G), i.e., G is Abelian.
2. |Z(G)| = p ⇒ [|G/Z(G)| = |G|/|Z(G)| = p²/p = p ] |G/Z(G)| = p ⇒ [Let p be a prime number, |G|=p ⇒ G is cyclic] G/Z(G) is cyclic ⇒ [Let G be a group and Z(G) be the center of G. If G/Z(G) is cyclic, then G is Abelian] G is Abelian.

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