Throughout is a group with the center and the commutator subgroup The goal is to prove that if is finite, then is finite too. We will also find an upper bound for in terms of

**Notation**. For any we define and

**Problem 1**. Let

1) and

2)

3) If then

*Proof*. 1) The first three identities are trivial and the fourth one is true because the map defined by is a group homomorphism.

2) By 1) we have

3) So for all because So with 1) give us

**Problem 2**. Let If then

*Proof*. Define the map by If we prove that is one-to-one, we are done because then So suppose that Then and and hence and Therefore

**Schur’s theorem**. If then

**Solution**. Let and Then where We will choose the integer to be as small as possible, i.e. if with then Now, we know from Problem 2, that the number of elements of is at most So if we prove that each can occur at most times in then and thus there will be at most possible values for and the problem is solved. To prove that each can occur at most times in suppose, to the contrary, that, say occurs times in the product. Then by part 2) of Problem 1, we can move each to the right hand side of the product to get where and But by part 3) of Problem 1, is a product of elements of and hence, since we see that is a product of elements of Therefore is a product of elements of which contradicts the minimality of

If is finitely generated, then the converse of Schur’s theorem is also true, i.e. if is finite, then is finite too. It’s not hard, try to prove it!