Derived subgroup where not every element is a commutator 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.
 A: Keith Dennis has also done some work on this problem.  In the paper:
Dennis, R. K.(1-CRNL); Vaserstein, L. N.(1-PAS)
Commutators in linear groups.
K-Theory 2 (1989), no. 6, 761–767. 
they consider conditions under which all elements of the commutator subgroup are products of at most 2 commutators.  I seem to recall that Dennis did computer calculations to find the smallest group for which the commutator subgroup contains elements that are not commutators (and my recollection is that the size of the group was around 100 elements) but I can't see anything on-line about that.
A: 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.
A: The "simplest" example is the free group on two generators, but is perhaps not the easiest to play with unless you are comfortable with free groups.
A nice survey of the problem of commutators vs. commutators subgroups in finite groups is:


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*Kappe, Luise-Charlotte, and Morse, Robert Fitzgerald. On commutators in groups. Groups St. Andrews 2005, Vol. 2, pp. 531-558. London Math. Soc. Lecture Notes Ser., 340, Cambridge University Press, 2007. MR 2331612 (2008g:20068)


The smallest groups in which the commutator subgroup does not equal the set of commutators have order 96; in fact, there are two non-isomorphic groups of order 96 in which the set of commutators does not equal the commutator subgroup. (This result also appeared in Robert Guralnick's 1977 thesis, "Expressing group elements as products of commutators", UCLA). 
Kappe also has an upcoming joint paper in which she discusses the same question, restricted to $p$-groups. 
Another interesting item is the Ore Conjecture, which stated that if $G$ is a finite nonabelian simple group, then the set of commutators equals the commutator subgroup. Its proof was recently completed, and it appears in:


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*Liebeck, Martin W.; O'Brien, E. A.; Shalev, Aner; Tiep, Pham Huu. The Ore conjecture. J. Eur. Math. Soc. (JEMS) 12 (2010), no. 4, 939–1008. MR MR2654085. A PDF preprint can be found at O'Brien's website.

A: An explicit example is $G = (V \times Q) \rtimes C$ where $V=\langle \varepsilon \rangle \times \langle \varepsilon \rangle \cong C_2 \times C_2$, $Q$ is the quaternion group $Q_8=\{\pm 1, \pm i, \pm j, \pm k\}$ and $C_3 \cong C =\langle t \rangle$ acts "diagonally" on both $V$ and $Q$. It acts on $V$ so that $VC \cong A_4$ and it acts on $Q$ by cyclically permuting $i,j,k$: $i \to j \to k \to i$, so that $QC \cong SL(\mathbb{F}_3^2)$.
We have $|G|=96$. It is a nice exercise to compute the character table of $G$, which is of size $12 \times 12$. If you do so, you will realize that the element $x=((1,\varepsilon),-1) \in V \times Q = G'$ satsifies $\sum_{i=1}^{12} \chi_i(x)/\chi_i(1)=0$ so it is not a commutator.
