FC not implies BFC

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This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., FC-group) need not satisfy the second group property (i.e., BFC-group)
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Statement

It is possible to have a FC-group G (i.e., every conjugacy class in G is finite) that is not a BFC-group (i.e., there is no finite upper bound on the sizes of conjugacy classes).

Proof

Proof

Suppose K is a finite group that is not abelian, so the derived subgroup of K is nontrivial. Let G be the restricted external direct product of a countably infinite number of copies of K. We can verify that:

  • G is a FC-group: any element of G has all but finitely many coordinates equal to the identity element of K, and therefore all its conjugates all have those coordinates as the identity element, so there can be only finitely many conjugates.
  • There is no finite upper bound on the sizes of conjugacy classes in G: Let x \in K be a non-central element, with a conjugacy class of size s > 1. Then, an element with n coordinates that are copies of x, and the remaining coordinates the identity element, has s^n conjugates. As n can be arbitrarily large, this allows for arbitrarily large conjugacy class sizes.