Commutator subgroup does not consist only of commutators?

The problem is whether the commutator subgroup may contain elements that are not commutators. One example are the free groups. For instance, in the free group of rank $4$, freely generated by $x$, $y$, $z$, and $w$, the element $[x,y][z,w]$ of the commutator subgroup cannot be written in the form $[a,b]$ for some $a,b$ in the group.

The smallest finite examples are groups of order 96; there's two of them, nonisomorphic to each other. (This was a result in Robert Guralnick's thesis). See this Math Stack exchange question for a description of these groups, and some references.

In $SL_2(\mathbb R)$ the element $-I$ is not a commutator. Proof: If $ABA^{-1}B^{-1}=-I$ then $ABA^{-1}=-B$, whence $B$ has trace $0$. Wlog, then, $B$ is a standard $90$ degree rotation matrix. Now $B^{-1}AB=-A$ forces $A$ to be such that its determinant has the form $-x^2-y^2$. Contradiction.

No-one has mentioned that which elements of the commutator subgroup are actually commutators can be determined from the character table, so I will. The commutator subgroup consists of those conjugacy classes in the kernel of every linear character of course, but it can also be shown that $g \in G$ is a commutator if and only if

$$ {\sum _\chi} \frac{\chi(g)}{ \chi(1)} \neq 0 $$

where the sum is over all irreducible characters $\chi$. This is exercise 3.10 in Isaacs' book on character theory. For the groups of order 96 it's reasonably practical to work out the character table (though not much fun).