Conjugacy Class

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A complete set of mutually conjugate group elements. Each element in a group belongs to exactly one class, and the identity element (I=1) is always in its own class. The conjugacy class orders of all classes must be integral factors of the group order of the group. From the last two statements, a group of prime order has one class for each element. More generally, in an Abelian group, each element is in a conjugacy class by itself.

Two operations belong to the same class when one may be replaced by the other in a new coordinate system which is accessible by a symmetry operation (Cotton 1990, p. 52). These sets correspond directly to the sets of equivalent operations.

To see how to compute conjugacy classes, consider the dihedral group D3, which has the following multiplication table.

D_31ABCDE
11ABCDE
AA1DEBC
BBE1DCA
CCDE1AB
DDCABE1
EEBCA1D

{1} is always in a conjugacy class of its own. To find another conjugacy class take some element, say A, and find the results of all similarity transformations X^(-1)AX=X^(-1)(AX) on A. For example, for X=A, the product of A by A can be read of as the element at the intersection of the row containing A (the first multiplicand) with the column containing A (the second multiplicand), giving A^(-1)AA=A^(-1)1. Now, we want to find Z where A^(-1)1=Z, so pre-multiply both sides by A to obtain (AA^(-1))1=1=AZ, so Z is the element whose column intersects row A in 1, i.e., A. Thus, A^(-1)AA=A. Similarly, B^(-1)AB=C, and continuing the process for all elements gives

A^(-1)AA=A
(1)
B^(-1)AB=C
(2)
C^(-1)AC=B
(3)
D^(-1)AD=C
(4)
E^(-1)AE=B.
(5)

The possible outcomes are A, B, or C, so {A,B,C} forms a conjugacy class. To find the next conjugacy class, take one of the elements not belonging to an existing class, say D. Applying a similarity transformation gives

A^(-1)DA=E
(6)
B^(-1)DB=E
(7)
C^(-1)DC=E
(8)
D^(-1)DD=D
(9)
E^(-1)DE=D,
(10)

so {D,E} form a conjugacy class.

Let G be a finite group of group order |G|, and let s be the number of conjugacy classes of G. If |G| is odd, then

 |G|=s (mod 16)
(11)

(Burnside 1955, p. 295). Furthermore, if every prime p_i dividing |G| satisfies p_i=1 (mod 4), then

 |G|=s (mod 32)
(12)

(Burnside 1955, p. 320). Poonen (1995) showed that if every prime p_i dividing |G| satisfies p_i=1 (mod m) for m>=2, then

 |G|=s (mod 2m^2).
(13)

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