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The class equation of is
Example
For , the conjugacy classes are
Thus the class equation is .
Facts & Assumptions
Given: The symmetric group on .
The class equation splits a finite group into its central singleton classes and its non-singleton conjugacy classes (The class equation for a finite group).
Conjugacy-class size is a centralizer index ( is a bijection, so whenever these cardinalities are finite).
The symmetric group consists of all permutations of the underlying set (The symmetric group : the bijections of a set under composition).
The symmetric group is a group under composition ( is a group under composition, and it is non-abelian whenever has at least three distinct elements).
A three-element set has bijections to itself (A finite set with has exactly bijections onto itself, and bijections onto any set of the same cardinality).
Verification
The identity, the three transpositions, and the two -cycles are six distinct permutations; by [L3] and [L5], they exhaust .
Conjugation relabels cycle entries: every transposition is conjugate to every other, and the two -cycles are conjugate. Cycle type is preserved by conjugation, so the three displayed sets are exactly the conjugacy classes.
Their cardinalities are , , and , so [L1] gives .
Depends on
- The class equation $|G|=|Z(G)|+\sum_i [G:C_G(x_i)]$ for a finite group
- $G/C_G(x)\to\operatorname{Cl}_G(x)$ is a bijection, so $|\operatorname{Cl}_G(x)|=[G:C_G(x)]$ whenever these cardinalities are finite
- The symmetric group $\operatorname{Sym}(X)$: the bijections of a set $X$ under composition
- $\operatorname{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements
- A finite set $A$ with $\lvert A\rvert = n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality
Used by
- S₃ acting on three points has |X|=3 and |X^S₃|=0, so the fixed-point congruence modulo 2 fails without the p-group hypothesis Counterexample
- S₃ has trivial center, so the finite p-group hypothesis in the nontrivial-center theorem is necessary Counterexample
- The three subgroups of order 2 in S₃ are conjugate and each is self-normalizing Example
Dependency tree · two levels
26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- T. W. Judson, Abstract Algebra: Theory and Applications, 14.2 (standard reference, not scraped)