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,
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.