Maximum number of commutators required to generate an element of the derived subgroup Let $G$ be a group for which the center $Z(G)$ is of index $n$.
How to prove that an element of the derived subgroup $G^\prime$ is the product of at most $n^3$ commutators?
This exercise is from a French algebra book of the 70s.
It follows another exercise asking to prove that if $G$ is a group whose center $Z(G)$ is of index $n$, then for any elements $x,y \in G$
$$[x,y]^{n+1}=[x,y^2][yxy^{-1},y]^{n-1}$$
 A: The first step is to prove that there are at most $n^2$ distinct commutators. Let $N = Z(G)$, and choose elements $g_1, \dots, g_n$, one in each coset of $G/N$, so $g_1N, \dots, g_nN$ form a partition of $G$. Then for any $x, y$, we have $x \in g_iN$ and $y \in g_jN$ for some $i, j$, hence there are some $h, k \in Z(G)$ such that $x = g_i h$ and $y = g_j k$. Using this, we have
$$[x, y] = x^{-1}y^{-1}xy = h^{-1} g_i^{-1} k^{-1} g_j^{-1} g_i h g_j k = h^{-1}hk^{-1}k g_i^{-1} g_j^{-1} g_i g_j = [g_i, g_j]$$
so every commutator is of the form $[g_i, g_j]$ for some $i, j$.
Now, let $g \in G'$, so $g$ can be expressed as a product of commutators, let $m$ be the minimum length of an expression of $g$ as a product of commutators, and let
$$g = c_1 c_2 \cdots c_m$$
be an expression of $g$ as a product of commutators of length $m$. Suppose $m > n^3$. Then at least one of the $n^2$ commutators, say $a$, appears in this product at least $n+1$ times, so we can write
$$g = h_0 a h_1 a h_2 a \cdots a h_n a h_{n+1}$$
where each $h_i$ is a (possibly empty) product of commutators. From here, the idea is that by "swapping" we can "drag" the $a$'s in this product to the front. Note that $a^{-1}[x, y]a = [a^{-1}xa, a^{-1}ya]$, so in particular, for any commutator $c$, there is another commutator $c'$ for which $ca = ac'$. Similarly, if $h$ is a product of $r$ commutators, then there is another $h'$ which is a product of $r$ commutators for which $ha = ah'$. Thus we can write
\begin{align*}
g 
&= h_0 a h_1 a h_2 a \cdots a h_n a h_{n+1} \\
&= a h_0' h_1 a h_2 a \cdots a h_n a h_{n+1} \\
&= a^2 h_0'' h_1' h_2 a \cdots a h_n a h_{n+1} \\
&\,\,\vdots \\
&= a^{n+1} h_0^* h_1^* h_2^* \cdots h_n^* h_{n+1}
\end{align*}
where at each step we have kept the length of the product constant. But assuming the previous exercise you mentioned, $a^{n+1}$ can be written as a product of $n$ commutators, so $g$ can be written as a product of $m-1$ commutators, contradicting minimality. Thus $m \leq n^3$.
