Given a rank2 group $G= < a,b> $ . Is it true and trivial that $ [G,G] = < [a,b], [b,a] > $ ?
Thanks !
Given a rank2 group $G= < a,b> $ . Is it true and trivial that $ [G,G] = < [a,b], [b,a] > $ ? Thanks ! 

closed as too localized by Fernando Muro, Chris Godsil, Igor Rivin, HJRW, Bugs Bunny Jul 10 '12 at 16:52This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question. 


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. 


Also, every finite nonAbelian 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 2generated, and the alternating groups are easily seen to be 2generated, 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. 


The commutator subgroup of the free group $\langle a,b \rangle$ is freely generated by the set $$\lbrace [a^n,b^m] \mid n,m \in \mathbb Z, nm \neq 0 \rbrace.$$ 


There is a very general theorem (see Proposition 4 in Chapter 1 of JP. Serre, Trees. SpringerVerlag 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. 

