Why is the set of commutators not a subgroup? I was surprised to see that one talks about the subgroup generated by the commutators, because I thought the commutators would form a subgroup. Some research told me that it's because commutators are not necessarily closed under product (books by Rotman and Mac Lane popped up in a google search telling me). However, I couldn't find an actual example of this. What is one? The books on google books made it seem like an actual example is hard to explain. 
Wikipedia did mention that the product $[a,b][c,d]$ on the free group on $a,b,c,d$ is an example. But why? I know this product is $aba^{-1}b^{-1}cdc^{-1}d^{-1}$, but why is that not a commutator in this group?

Here is an explicit example, paraphrased from the paper of Martin Isaacs linked in the comments. Let $S_3$ be the group of permutations of a 3-element set, and consider the group ring $\mathbb Z_2[S_3]$, which is an abelian group under addition and admits an action of $S_3$ by right multiplication. Thus we can construct the semidirect product $G=S_3\ltimes\mathbb Z_2[S_3]$. Explicitly the operation is
$$
  (w_1,x_1)(w_2,x_2)=(w_1w_2,x_1w_2+x_2).
$$
We have
$$\begin{eqnarray*}
  [(w_1,x_1),(w_2,x_2)]
    &=&(w_1,x_1)(w_2,x_2)(w_1^{-1},x_1w_1^{-1})(w_2^{-1},x_2w_2^{-1})\\
    &=&([w_1,w_2],x_1(w_2+1)w_1^{-1}w_2^{-1}+x_2(w_1^{-1}+1)w_2^{-1})
\end{eqnarray*}$$
For any $w_1,w_2\in S_3$, the element
$$
  (1,w_1+w_2)=[(1,w_1),(w_2^{-1}w_1,0)]
$$
is a commutator. For $S\subseteq S_3$, let $\bar S=\sum_{s\in S}s\in\mathbb Z_2[S_3]$. Let $T=\{1,(123)\}$. Then
$$
  (1,\bar G+\bar T)=(1,(12)+(13))(1,(23)+(321))
$$
is a product of commutators. Suppose it equals $[(w_1,x_1),(w_2,x_2)]$. Then $[w_1,w_2]=1$ and
$$
  \bar G+\bar T=x_1(w_2+1)w_1^{-1}w_2^{-1}+x_2(w_1^{-1}+1)w_2^{-1}.
$$
Now $w_1,w_2$ generate an abelian subgroup $H<S_3$, and $w_1\bar H=w_2\bar H=\bar H$. Thus
$$
  (\bar G+\bar T)\bar H=0.
$$
By considering each possibility for $H$, we see that this is impossible.
