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 $G$ be a group. If the center $Z(G)$ is a finite-index subgroup of $G$, then the commutator subgroup $D(G)$ 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 $[x,y] = x y x^{-1} y^{-1}$.

Lemma 1: For all $a,x,y \in G$, $a[x,y]a^{-1} = [axa^{-1},aya^{-1}]$.

Lemma 2: If $Z(G)$ is a subgroup of finite index $n$, then $[x,y]^{n+1} = [x,y^2] \cdot [yxy^{-1}, y ]^{n-1}$.

Proof. Notice that, because $Z(G)$ is a subgroup of finite index $n$, for all $a \in G$ we have $a^n \in Z(G)$. Therefore,

$[x,y]^{n+1} = [x,y]^n \cdot xyx^{-1} y^{-1} = xyx^{-1} \cdot [x,y]^n \cdot y^{-1}$.

Then,

$[x,y]^{n+1} = xyx^{-1} \cdot [x,y] \cdot [x,y]^{n-1} \cdot y^{-1} = xy^2 x^{-1} y^{-2} \cdot y [x,y]^{n-1} y^{-1}$.

Using Lemma 1, we conclude that

$[x,y]^{n+1} = [x,y^2] \cdot \left( y[x,y]y^{-1} \right)^{n-1} = [x,y^2] \cdot [yxy^{-1},y]^{n-1}$. $\square$

Proof of Theorem: Let us suppose that $Z(G)$ is a subgroup of finite index $n$.

It is not difficult to notice that $[x,y]=[a,b]$ in $G$ whenever $x=a$ and $y=b$ in $G/Z(G)$. Therefore, $G$ has at most $n^2$ commutators.

Now, any $c \in D(G)$ can be written as a product of commutators

$c = c_1 \cdot c_2 \cdots c_r$.

Suppose that $r$ is as small as possible, and assume by contradiction that $r>n^3$. Therefore, because there exist at most $n^2$ commutators, the product contains some commutator, say $k$, at least $(n+1)$ times. We claim that it is possible to write

$c = k^{n+1} \cdot c_1' \cdots c_{r-n-1}' \ \ \ \ \ \ (1)$

for some new commutators $c_j'$. For example, if $c_i=k$, notice that

$c = c_1 \cdots c_{i-1} \cdot k \cdot c_{i+1} \cdots c_r = k \cdot (k^{-1}c_1k) \cdots (k^{-1}c_{i-1}k) \cdot c_{i+1} \cdots c_r$.

Thanks to Lemma 1, we know that each $k^{-1}c_jk$ is again a commutator. We just proved our claim for $n=0$. The general case follows by induction.

From $(1)$ and Lemma 2, we deduce that $c$ can be written as a product of $r-1$ commutators, a contradiction with the minimality of $r$.

Therefore, we just proved that any element of $D(G)$ can be written as a product of at most $n^3$ commutators. Because there exist at most $n^2$ commutators, we deduce that the cardinality of $D(G)$ is bounded above by $n^{2n^3}$; in particular, $D(G)$ is finite. $\square$

Corollary 1: If $G$ has only finitely many commutators, then $D(G)$ is finite.

Proof. Let $[g_1,g_2], \ldots, [g_{2r-1},g_{2r}]$ denote the commutators of $G$ and $H$ be the subgroup generated by the $g_i$‘s. Notice that $D(G)=D(H)$, so it is sufficient to prove that $D(H)$ is finite to conclude. Furthermore, according to our previous theorem, it is sufficient to prove that $Z(H)$ is a finite-index subgroup of $H$.

If $C(g_i)$ denotes the centralizer of $g_i$ in $H$, notice that

$Z(H)= \bigcap\limits_{i=1}^{2r} C(g_i)$.

Therefore, it is sufficient to prove that each $C(g_i)$ is a finite-index subgroup of $H$. But $H/C(g_i)$ is finite if and only if $g_i$ has only finitely-many conjugacy classes. Because $hg_ih^{-1}=[h,g_i] g_i$ for all $h \in H$, and because $H$ has only finitely-many commutators, it is clear that $g_i$ has finitely-many conjugacy classes, concluding our proof. $\square$

Corollary 2: The only infinite group all of whose non-trivial subgroups are finite-index is $\mathbb{Z}$.

Proof. Let $G$ be such a group. Let $g_0 \in G \backslash \{ 1 \}$ and $g_1, \ldots, g_r \in G \backslash \{ 1 \}$ be a set of representatives of the cosets of $\langle g_0 \rangle$. For each $0 \leq i \leq r$, $\langle g_i \rangle$ is a finite-index subgroup of $G$ by assumption; but it is also a subgroup of the centralizer $C(g_i)$, so $C(g_i)$ is a finite-index subgroup of $G$. Therefore, because $\{ g_0, \ldots, g_r \}$ generates $G$, the center

$Z(G)= \bigcap\limits_{i=0}^r C(g_i)$

is a finite-index subgroup of $G$. According to our previous theorem, $D(G)$ is finite, and by assumption, any subgroup of $G$ is either trivial, or finite-index and in particular infinite since $G$ so is. Therefore, $D(G)$ is trivial, that is $G$ 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 $\mathbb{Z}$. $\square$

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 $D_{\infty}$ is such an example.

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

Proof. Let $G$ be a torsion-free group with an infinite cyclic subgroup $H= \langle g_0 \rangle$ of finite index $n$. As above, if $g_1, \ldots, g_r$ denotes a set of representatives of the cosets of $\langle g_0 \rangle$ then $\{ g_i \mid 0 \leq i \leq r \}$ generates $G$. Noticing that $g_i^n \in H$, we deduce that $C(g_i) \cap H$ is of finite index in $H$, so the centralizer $C(g_i)$ has a finite index in $G$. Again, we deduce that the center

$Z(G)= \bigcap\limits_{i=0}^r C(g_i)$

is a finite-index subgroup of $G$. According to our previous theorem, $D(G)$ is finite, and in fact trivial because $G$ is torsion-free, so $G$ 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 $\mathbb{Z}$. $\square$

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 $G$ is abelian if and only if $Z(G)=G$ if and only if $D(G)= \{ 1 \}$. Therefore, a way to say that $G$ is almost abelian is to state that $Z(G)$ is big (eg. is a finite-index subgroup) or that $D(G)$ 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 $G$ be a finitely-generated group. If $D(G)$ is finite then $Z(G)$ is a finite-index subgroup.

Proof. Let $g_1, \ldots, g_r \in G$ be a finite generating set of $G$. Then

$Z(G)= \bigcap\limits_{i=1}^r C(g_i)$.

To conclude, it is sufficient to prove that each $C(g_i)$ is a finite-index subgroup, that is equivalent to say that each $g_i$ has finitely-many conjugacy classes. Noticing that for all $h \in G$, $hg_ih^{-1}=[h,g_i]g_i$, we deduce the latter assertion because $D(G)$ is finite. $\square$