    # 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(Ro) = {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, ···, ak 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(ai)|·|C(ai)|, then |G|/|C(ai)| = |cl(ai)| = |G:C(ai)|

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

G = cl(b1) ∪ ··· cl(br) ∪ cl(a1) ∪ ··· cl(ak) ⇒ [A partition is a union of disjoint classes] |G| = |cl(b1)| + ··· + |cl(br)| + |cl(a1)| + ··· + |cl(ak)| = [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(ak)| = |Z(G)| + |G:C(a1)| + ··· + |G:C(ak)| ⇒ $|G| = |Z(G)|+\sum_{i=1}^k |G:C(a_k)|$

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 = gn1i1 and h2 = gn2i2

h1h2 = gn1i1gn2i2 =[i1 ∈ Z(G)] gn1gn2i1i2 = gn1+n2i1i2 = gn2gn1i1i2 =[i2 ∈ Z(G)] gn2gn1i2i1 =[i2 ∈ Z(G)] gn2i2gn1i1 = 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(ai)|=|G|/|C(ai)| where C(ai) is a subgroup, its order must divide |G|, |G|/|C(ai)| = $\frac{p^n}{p^k}$, 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.

# 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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