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Given a rank-2 group $G= < a,b> $ . Is it true and trivial that $ [G,G] = < [a,b], [b,a] > $ ?

Thanks !

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closed as too localized by Fernando Muro, Chris Godsil, Igor Rivin, HJRW, Bugs Bunny Jul 10 '12 at 16:52

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By the way, $[b,a]=[a,b]^{-1}$ so your question asks whether the derived subgroup of any 2-generated group is cyclic. – Mark Sapir Jul 9 '12 at 8:03
Out of interest, where are all these questions on group theory coming from? – Yemon Choi Jul 9 '12 at 8:38
I find it difficult to believe that this trivial question needs four distinct answers. – HJRW Jul 10 '12 at 11:18
up vote 5 down vote accepted

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.

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Thanks a lot !!!! – jason mfash Jul 9 '12 at 8:14

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.

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The commutator subgroup of <a,b> is of course the normal subgroup generated by [a,b], that is the subgroup generated by all conjugates of [a,b]. – Geoff Robinson Jul 9 '12 at 21:44

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

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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.

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