Is there a better bound than $(n^2)^{n^3}$ for the order of the commutator subgroup of a group whose center has index $n?$

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.

Here's an argument for $(n+1)^{n^2}$. I'll try and be a bit formal, at the risk of making it sound mote complicated than it is. Let $c_1,\cdots,c_k$ be an enumeration of the set $\{c_{11},\cdots,c_{nn}\}$. We will write an arbitrary element $g$ of the commutator subgroup in the form $c_1^{p_1}\cdots c_k^{p_k}$ with $0\leq p_1,\cdots,p_k \leq n$.

There exist $p_1,\cdots,p_k\geq 0$ and a sequence $i_1,\cdots,i_m$ such that $g=c_1^{p_1}\cdots c_k^{p_k}c_{i_1}\cdots c_{i_m}$ (for example with $p_1=\cdots=p_k=0$). So we can pick such a representation of $g$ such that $p_1+\cdots+p_k+m$ is minimal, and then fixing that, such that $p_1$ is maximal, and then such that $p_2$ is maximal and so on. Suppose for contradiction that $m>0$. Then we can move the $c_{i_1}$ term back using rule 1 to get a representation of $g$ such that $p_{i_1}$ is larger, which is a contradiction. Suppose for contradiction that $p_i>n$. Then we can use rule 2 and get a contradiction to the minimality of $p_1+\cdots+p_k+m$.

Another small improvement is $(2n+1)^{n(n-1)/2}$ by letting $-n\leq p_i\leq n$ and using $[x,y]=[y,x]^{-1}$.

It's been 7 years since the question was posted, but now I found myself here and I know better bounds, so I'm answering anyway. The simple proof uses the well-known fact that a finite group $G$ is always generated by a subset of $G$ order less or equal than $[log_2|G|]$.

Let $G$ be a group, let $T$ be a (right) transversal of $Z(G)$ in $G$, which has order $n$ by hypothesis, and let $H=\langle T\rangle$. Since clearly $G'=H'$, we can assume that $G$ is generated by $[log_2n]$ elements. Now, by the Reidemeister-Schreier Theorem $G'\cap Z(G)$ can be generated by $n[log_2n]-n+1$ elements and has hence at worst order $n^{n[log_2n]-n+1}$ since it is a finitely generated abelian group of exponent $n$. It follows that $$|G'|\leq n^{n[log_2n]-n+2}$$

Remark: In Multiplicators and groups with finite central factor-groups. Math. Z. 89, 345-347 (1965), James Wiegold provides with even better bounds.