FZ implies finite derived subgroup

This result was proved by Schur and is sometimes termed the Schur-Baer theorem.

Verbal statement
If the inner automorphism group (viz the quotient by the center) of a group is finite, so is the derived subgroup. In fact, there is an explicit bound on the size of the commutator subgroup as a function of the size of the inner automorphism group.

Symbolic statement
Let $$G$$ be a group such that $$\operatorname{Inn}(G) = G/Z(G)$$ is finite. Then, $$G' = [G,G]$$ is also finite. In fact, if $$|G/Z(G)| = n$$, then $$G'$$ has size at most $$n^{2n^3}$$.

Variety-theoretic statement
The variety of abelian groups is a Schur-Baer variety.

Other facts about related group properties

 * Finitely many commutators implies finite derived subgroup: If the number of elements in a group that can be written as commutators is finite, then the derived subgroup is finite. Note that this fact is proved using the Schur-Baer theorem, hence it cannot be used to simplify the proof of the Schur-Baer theorem given below.
 * Finitely generated and FC implies FZ

Alternative formulation

 * Exterior square of finite group is finite

Facts used

 * 1) uses::Exterior square of finite group is finite
 * 2) uses::Commutator map is homomorphism from exterior square to derived subgroup of central extension (and this homomorphism is surjective by definition)

Proof using the alternate formulation
Given: A group $$G$$ such that $$\operatorname{Inn}(G)$$ is finite.

To prove: $$G'$$ is finite and its order is bounded in terms of the order of $$\operatorname{Inn}(G)$$.

Proof:

Direct proof outline
The proof involves two steps:


 * Showing that the number of distinct commutators is at most $$n^2$$: This follows from the fact that the commutator $$[x,y]$$ depends only on the quotients of $$x$$ and $$y$$ modulo $$Z(G)$$, and thus there are $$n^2$$ possibilities.
 * Showing that for any element in the commutator subgroup, there is a minimal word for that element with each commutator occuring at most $$n$$ times: This shows that any element of the commutator subgroup has a word in terms of the commutators, of length at most $$n^3$$, and this completes the proof.