TITLE: Commutator subgroup does not consist only of commutators?
QUESTION [25 upvotes]: Let $G$ be a group, $G'=[G, G]$.

"Note that it is not necessarily true that the commutator subgroup 
    $G'$ of $G$ consists  entirely   of
    commutators $[x, y], x, y \in G$ (see [107] for some finite group examples)." 

Quoted from http://www.math.ucdavis.edu/~kapovich/EPR/ggt.pdf page 8.
Anybody can provide the examples? I can't find the book [107].

REPLY [4 votes]: Sorry for adding one more answer, but here's a simple argument based on using that if the center is large enough then the commutator map has too small image to cover the derived subgroup, and even showing that, in suitable (varying) finite groups, the commutator length can be unbounded.
Let $p$ be an odd prime. Consider the free 2-step-nilpotent Lie algebra over $\mathbf{Z}/p\mathbf{Z}$ on $n$ generators. It has dimension $n+n(n-1)/2$, and its center has dimension $n(n-1)/2$ and coincides with the derived subalgebra.
The law $x\ast y=x+y+\frac12[x,y]$ defines a group law (this is the Baker-Campbell-Hausdorff formula), defining a group $G$ of order $p^{n+n(n-1)/2}$  (and exponent $p$). 
The center $Z$ of $G$ has order $p^{n(n-1)/2}$ and coincides with the derived subgroup. The commutator map $G\times G\to [G,G]$ factors through $G/Z\times G/Z$, and hence its image contains at most $p^{2n}=|G/Z|^2$ elements. As soon as $2n<n+n(n-1)/2$ (i.e., $n\ge 4$), this implies that there are non-commutators.
Better, this shows that the set of products of $k$ commutators has cardinal $\le p^{2nk}$, and therefore does not cover the commutator group if $2nk<n+n(n-1)/2$, that is, $n\ge 4k$.
Notes:
1) This example (at least the existence of product of commutators that are not commutators) is certainly covered by Torsten's answer; however I provide a different argument.
2) More complicated elaborations, also based on the center being large, yield finite perfect groups with arbitrary large commutator width.
3)One can directly define $G$ as the relatively free group on $n$ generators in the variety of 2-step-nilpotent groups of exponent $p$, but then one needs to check by hand the statement on the derived subgroup and center.