Size of conjugacy class divides index of center

This article gives the statement, and possibly proof, of a constraint on numerical invariants that can be associated with a finite group

This article states a result of the form that one natural number divides another. Specifically, the (size) of a/an/the (conjugacy class) divides the (center) of a/an/the (index of a subgroup).
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Statement

Statement with symbols

Suppose $G$ is a group and $Z(G)$ is its center. Suppose further that the index $[G:Z(G)]$ is finite. Let $K$ be a conjugacy class in $G$; in other words, $K$ is an orbit under the action of the group on itself by conjugation. We then have that $K$ is finite, and further:

$|K| | [G:Z(G)]$.

Related facts

Stronger facts

• Size of conjugacy class equals index of centralizer

Weaker facts

• Size of conjugacy class divides order of group

Other facts about size of conjugacy class

• Size of conjugacy class is bounded by order of derived subgroup
• Size of conjugacy class need not divide order of derived subgroup
• Size of conjugacy class need not divide index of abelian normal subgroup

Similar facts about degrees of irreducible representations

• Degree of irreducible representation divides index of center
• Degree of irreducible representation divides index of abelian normal subgroup
• Degree of irreducible representation is bounded by index of abelian subgroup

Facts about groups in which the index of the center is finite

A group where the center has finite index is termed a FZ-group, while a group where the conjugacy class of every element is finite in size is termed a FC-group. This result implies that every FZ-group is a FC-group.

FZ-groups have a number of interesting properties. For instance, the Schur-Baer theorem asserts that the derived subgroup of a FZ-group is finite. In fact, if the center has index $n$, the order of the commutator subgroup is bounded by $n^{2n^3}$.

Other related facts

• Lagrange's theorem: This states that the order of any subgroup divides the order of the group.
• Fundamental theorem of group actions: This states that for any group action, there is a bijection between the orbit of an element and the left coset space of its stabilizer. In particular, it shows that the size of any orbit divides the order of the group.
• Order of quotient group divides order of group: This is a consequence of Lagrange's theorem and the first isomorphism theorem.
• Degree of irreducible representation divides group order
• Degree of irreducible representation divides index of center
• Degree of irreducible representation divides index of abelian normal subgroup

Facts used

1. Left coset space of centralizer is in bijective correspondence with conjugacy class
2. Index is multiplicative: If $A \le B \le C$ are groups, then $[C:A] = [C:B][B:A]$.

Proof

Given: A finite group $G$ with center $Z(G)$. $[G:Z(G)]$ is finite. A conjugacy class $K$ in $G$.

To prove: $K$ is finite and the size of $K$ divides $[G:Z(G)]$.

Proof:

Step no.	Assertion/construction	Facts used	Given data/assumptions used	Previous steps used	Explanation
1	Let $g \in K$ and $L = C_G(g)$. Then, $L$ contains $Z(G)$				$Z(G)$ is the set of elements that commute with every element, therefore, it is contained in $C_G(g) = L$.
2	$L$ is a subgroup of finite index in $G$ and $[G:Z(G)] = [G:L][L:Z(G)]$, so in particular the index of $L$ divides $[G:Z(G)]$	Fact (2)		Step (1)	Step-fact combination direct.
3	$|K|$ is finite and divides $[G:Z(G)]$	Fact (1)		Step (2)	By Fact (1), $|K| = [G:L]$. By Step (2), the latter divides $[G:Z(G)]$, hence so does the former.

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