Let $G$ be a group and let $G'$ be the derived subgroup, defined as the subgroup generated by the commutators of $G$.
Is there an example of a finite group $G$ where not every element of $G'$ is a commutator? $G'$ is only generated by commutators, but with all of the properties of commutators (ie: what happens under conjugation, exponentation, etc) I can't think of an example.
Best Answer
For any prime $p$ and $n>1$, there are nilpotent groups $G$ of class 2 and order $p^{n(n+1)/2}$ with generators $a_i$ $(1 \le i \le n)$, $b_{ij}$ $(1 \le i < j \le n)$, such that $[a_i,a_j] = b_{ij}$, the $b_{ij}$ are all central in the group, and all generators have order $p$.
Then $G'$ is the group of order $p^{n(n-1)/2}$ generated by the $b_{ij}$.
In any group, we have $[ax,by] = [a,b]$ when $x,y$ are central in the group, so $G$ has at most $p^{2n}$ distinct elements that are commutators.
Hence, for any fixed $k>0$, by choosing $n$ sufficiently large we can find $G$ such that not all element of $G'$ are products of at most $k$ commutators.