Let $G$ be a group and $[G:Z(G)]=n<\infty,$ where $Z(G)$ is the center of $G$. A theorem says that in this case
$o([G,G])\leq(n^2)^{n^3}, $
where $o(\cdot)$ denotes order and $[G,G]=G'$ is the commutator subgroup of $G$.
The bounds from the proof seem crude and I would like to ask if we can improve upon the result. Of course, the theorem is valuable with any bound because it implies that $G'$ is finite.
Proof. We have exactly $n$ cosets $\{x_iZ(G)\}_{i=1}^n$. We will be using the representatives $x_i.$ Let $c_{ij}=[x_i,x_j]=x_i^{-1}x_j^{-1}x_ix_j.$
First, we note the following
Lemma 1.
$\{c_{ij}\,|\,i,j=1,...,n\}=\{[x,y]\,|\,x,y\in G\}.$
Proof of lemma. For $x,y\in G$ we have some $i,j\in\{1,...,n\}$ and $z_1,z_2\in Z(G)$ such that
$x=x_iz_1\text{ and }y=x_jz_2.$
Then
$ \begin{eqnarray} [x,y]&=&x^{-1}y^{-1}xy=z_1^{-1}x_i^{-1}z_2^{-1}x_j^{-1}x_iz_1x_jz_2\\ &=& (x_i^{-1}x_j^{-1}x_ix_j)(z_1^{-1}z_1z_2^{-1}z_2)\\ &=& x_i^{-1}x_j^{-1}x_ix_j\\ &=& [x_i,x_j]. \end{eqnarray} $
Lemma 2. For $x,y\in G$ we have
$[x,y]^{n+1}=[x,y^2][y^{-1}xy,y]^{n-1}.$
Proof of lemma. Since $G/Z(G)$ has order $n$, we have $[x,y]^n\in Z(G)$. Therefore
$ \begin{eqnarray} [x,y]^{n+1}&=&x^{-1}y^{-1}xy[x,y]^n\\ &=& x^{-1}y^{-1}x[x,y]^ny\\ &=& x^{-1}y^{-1}x(x^{-1}y^{-1}xy)[x,y]^{n-1}y\\ &=& (x^{-1}y^{-2}xy^2)y^{-1}[x,y]^{n-1}y\\ &=& [x,y^2][y^{-1}xy,y^{-1}yy]^{n-1}\\ &=& [x,y^2][y^{-1}xy,y]^{n-1} \end{eqnarray} $
Proof cont'd. Let $g\in G'.$ It is a product of a a finite number $m$ of elements $c_{ij}$ (with possible repetitions) because $[x,y]^{-1}=[y,x]$ and so we do not need to consider inverses. Suppose $m>n^3$. We have
$\operatorname{card}(\{c_{ij}\,|\,i,j=1,...,n\})\leq n^2$
so some element $c_{ij},$ say $c=[x,y],$ must appear at least $n+1$ times in the product. There is no reason to believe however that $c^{n+1}$ occurs in the product since the occurances of $c$ may be scattered. We fix it by noting that for any $x',y'\in G$ we have
$[x',y'][x,y]=[x,y]c^{-1}[x',y']c=[x,y][c^{-1}x'c,c^{-1}yc].\tag1$
This allows us to rewrite $g$ as a product of commutators which begins with $[x,y]^{n+1}:$
$g=[x,y]^{n+1}c_{n+1}c_{n+1}...c_{m}.$
But this equals
$[x,y^2][y^{-1}xy,y]^{n-1}c_{n+1}c_{n+1}...c_{m},$
which is a product of $m-1$ commutators. Repeating this procedure, we can decrease the number of factors to $n^3.$ Therefore we can write any element of $G'$ as a product of at most $n^3$ commutators. We recall that there can be at most $n^2$ distinct commutators in $G$ and therefore there are at most $(n^2)^{n^3}$ elements in $G'.$
EDIT I have laid my hands on a copy of Passman's book on group rings and I see that the theorem and the proof come from it. The author attributes the theorem to Schur.