# 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

### 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 $n^{2n^3}$.

## 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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