The aim of this note is to prove the following theorem, due to Schur, and to exhibit some corollaries. The proof given here comes from J. Dixon’s book, *Problems in Group Theory* (problems 5.21 to 5.24).

**Theorem: **Let be a group. If the center is a finite-index subgroup of , then the commutator subgroup is finite.

We begin with two easy lemmae. The first one is left to the reader as an exercice: it is just a computation. In the sequel, we use the notation .

**Lemma 1:** For all , .

**Lemma 2:** If is a subgroup of finite index , then .

**Proof.** Notice that, because is a subgroup of finite index , for all we have . Therefore,

.

Then,

.

Using Lemma 1, we conclude that

.

**Proof of Theorem:** Let us suppose that is a subgroup of finite index .

It is not difficult to notice that in whenever and in . Therefore, has at most commutators.

Now, any can be written as a product of commutators

.

Suppose that is as small as possible, and assume by contradiction that . Therefore, because there exist at most commutators, the product contains some commutator, say , at least times. We claim that it is possible to write

for some new commutators . For example, if , notice that

.

Thanks to Lemma 1, we know that each is again a commutator. We just proved our claim for . The general case follows by induction.

From and Lemma 2, we deduce that can be written as a product of commutators, a contradiction with the minimality of .

Therefore, we just proved that any element of can be written as a product of at most commutators. Because there exist at most commutators, we deduce that the cardinality of is bounded above by ; in particular, is finite.

**Corollary 1:** If has only finitely many commutators, then is finite.

**Proof.** Let denote the commutators of and be the subgroup generated by the ‘s. Notice that , so it is sufficient to prove that is finite to conclude. Furthermore, according to our previous theorem, it is sufficient to prove that is a finite-index subgroup of .

If denotes the centralizer of in , notice that

.

Therefore, it is sufficient to prove that each is a finite-index subgroup of . But is finite if and only if has only finitely-many conjugacy classes. Because for all , and because has only finitely-many commutators, it is clear that has finitely-many conjugacy classes, concluding our proof.

**Corollary 2:** The only infinite group all of whose non-trivial subgroups are finite-index is .

**Proof.** Let be such a group. Let and be a set of representatives of the cosets of . For each , is a finite-index subgroup of by assumption; but it is also a subgroup of the centralizer , so is a finite-index subgroup of . Therefore, because generates , the center

is a finite-index subgroup of . According to our previous theorem, is finite, and by assumption, any subgroup of is either trivial, or finite-index and in particular infinite since so is. Therefore, is trivial, that is is abelian.

From the classification of finitely-generated abelian groups, it is not difficult to prove that the only abelian group all of whose non-trivial subgroups are finite-index is .

Notice however that there exist non-cyclic groups all of whose normal subgroups have finite-index: they are called *just-infinite groups*. The infinite dihedral group is such an example.

**Corollary 3:** A torsion-free virtually infinite cyclic group is infinite cyclic.

**Proof.** Let be a torsion-free group with an infinite cyclic subgroup of finite index . As above, if denotes a set of representatives of the cosets of then generates . Noticing that , we deduce that is of finite index in , so the centralizer has a finite index in . Again, we deduce that the center

is a finite-index subgroup of . According to our previous theorem, is finite, and in fact trivial because is torsion-free, so is abelian. From the classification of finitely-generated abelian groups, it is not difficult to prove that the only abelian torsion-free virtually cyclic group is .

In fact, using Stallings theorem, it is possible to prove more generally that any torsion-free virtually free group turns out to be free.

A group is abelian if and only if if and only if . Therefore, a way to say that is *almost* abelian is to state that is *big* (eg. is a finite-index subgroup) or that is *small* (eg. is finite). Essentially, our main theorem proves that the first idea implies the second. Although the converse does not hold in general, it holds for finitely-generated groups:

**Theorem:** Let be a finitely-generated group. If is finite then is a finite-index subgroup.

**Proof.** Let be a finite generating set of . Then

.

To conclude, it is sufficient to prove that each is a finite-index subgroup, that is equivalent to say that each has finitely-many conjugacy classes. Noticing that for all , , we deduce the latter assertion because is finite.