Commutator Subgroup - Group Theory Given a rank-2 group $G= < a,b> $ . 
Is it true and trivial that $ [G,G] = < [a,b], [b,a] > $ ? 
Thanks ! 
 A: No, the derived subgroup of the free group of rank 2 is infinitely generated (as every normal subgroup of infinite index), see W. Magnus, A. Karras, D. Solitar, Combinatorial group theory.
A: Also, every finite non-Abelian simple group is generated by two elements, and no such group has cyclic derived subgroup. (R. Steinberg proved in a uniform manner that simple groups of Lie type are 2-generated, and the alternating groups are easily seen to be 2-generated, so it isn't necessary to invoke the full classification of finite simple groups to answer the question in the negative). For that matter, for $n >3,$ the symmetric group $S_{n} = \langle (12),(12....n) \rangle,$ but its derived group is $A_{n},$ which is not cyclic.
A: The commutator subgroup of the free group $\langle a,b \rangle$ is freely generated by the set
$$\lbrace [a^n,b^m] \mid n,m \in \mathbb Z, nm \neq 0 \rbrace.$$
A: There is a very general theorem (see Proposition 4 in Chapter 1 of J-P. Serre, Trees. Springer-Verlag Berlin Heidelberg (1980).) which says: Let $A$ and $B$ be two groups. The kernel of the natural quotient map $A\ast  B\rightarrow A\times B$ is a free group generated by all commutators of the form $[a,b]$ where $a\in A - \{1\}$ and $b\in B - \{1\}$. Your question is the special case that $A=B=\mathbb{Z}$. In short, it coincides exactly with Andreas' answer. 
