Let $A,B$ be two generic (in particular invertible) $2\times 2$ upper-triangular complex matrices. They generate a countable group $G$, the commutator subgroup of $G$ is abelian. Are there other relations in $G$? How is this group called?
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asked Dec 5, 2019 at 7:24
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$\begingroup$ What do you call "relation" in a group? "The commutant of any two generic element is abelian" is not usually called a relation. $\endgroup$– YCorCommented Dec 5, 2019 at 8:18
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$\begingroup$ any two commutators commute, it is kind of relation, is not it? $\endgroup$– Fedor PetrovCommented Dec 5, 2019 at 8:20
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$\begingroup$ Oh, what you call the commutant is known as the derived subgroup (or you mean the set of commutators), but "commutant" means something else, namely centralizer (Wikipedia bicommutant). $\endgroup$– YCorCommented Dec 5, 2019 at 8:42
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$\begingroup$ You need to fix the size $k$ of matrices. The group is, I think, isomorphic to the (relatively) free group on two generators in the group variety (in the sense of universal algebra) generated by the group of $k\times k$ upper triangular matrices (over $\mathbf{C}$, but this variety does depend on the choice of field of characteristic zero). For instance for $k=1$ this is a free abelian group in 2 generators and for $k=2$ this is a free metabelian group on 2 generators (but for $k=3$ it will not be free 3-solvable on 2 generators as the latter is not linear). $\endgroup$– YCorCommented Dec 5, 2019 at 8:46
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$\begingroup$ Oops, I just noticed you stick to $k=2$. Then you get the free metabelian group on 2 generators. I'll post an answer. $\endgroup$– YCorCommented Dec 5, 2019 at 8:48
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1 Answer
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I think it is a old result of Magnus that the free metabelian group on $d$ generators can be embedded as group $T_2$ of $2\times 2$ matrices over $\mathbf{C}$.
It is easy to deduce that
every generic $d$-tuple in $T_2$ freely generates such a free metabelian group $\Gamma_d$.
Here generic means: not in some countable union of proper Zariski closed subsets of $T_2^d$; hence it's generic both in the topological and the measured sense.
Indeed, let $w(x_1,\dots,x_d)$ be a nontrivial element $\Gamma_d$. Let $F_w$ be the set of $d$-tuples $g\in T_2^d$ such that $w(g)=I_2$. Then $F_w$ is a proper Zariski closed subset. Hence any $d$-tuple outside $\bigcup_{w\in\Gamma_d-\{1\}}F_w$ works.
(A group is metabelian if its derived subgroup is abelian, or equivalently if and only if any two commutators commute. So it doesn't mean a free metabelian group satisfies no other relation, but it indeed means that all its relations can be derived from this one.)
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$\begingroup$ thank you! of course by "other relations" I meant "relations not implied by..." $\endgroup$ Commented Dec 5, 2019 at 10:21