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no, it is not true. the following is already contained in Andreas Thom question.

from the first paragraph of his question:

Let $n$ be an integer. Camille Jordan showed that there exists some $m \in {\mathbb N}$ (depending on $n$), such that for any pair of $n \times n$-unitaries $u,v \in U(n)$ which generate a finite group, one has $[u^m,v^m] = 1_n$.(In fact, he showed that any finite subgroup of $GL(n)$ has a abelian normal subgroup of finite index, bounded independently of the subgroup.)

Take $u_1,u_2\in{U}(n)$ that generate a free group (easy to construct for $n\geq{2}$), and let $m$ be as above. Then, since $v_1^m,v_2^m$ commute, $$\|u_1^mu_2^m-u_2^mu_1^m\|\leq{}2\|u_1^m-v_1^m\|+2\|u_2^m-v_2^m\|\leq{}4m\varepsilon$$

Since $\epsilon$ is arbitrary we have a contradiction.

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no, it is not true. the following is already contained in Andreas Thom question.

from the first paragraph of his question:

Let $n$ be an integer. Camille Jordan showed that there exists some $m \in {\mathbb N}$ (depending on $n$), such that for any pair of $n \times n$-unitaries $u,v \in U(n)$ which generate a finite group, one has $[u^m,v^m] = 1_n$. (In fact, he showed that any finite subgroup of $GL(n)$ has a abelian normal subgroup of finite index, bounded independently of the subgroup.)

Take $u_1,u_2\in{U}(n)$ that generate a free group (easy to construct for $d\geq{2}$), n\geq{2}$), and let$m$be as above. Then, since$v_1^m,v_2^m$commute, $$\|u_1^mu_2^m-u_2^mu_1^m\|\leq{}2\|u_1^m-v_1^m\|+2\|u_2^m-v_2^m\|\leq{}4m\varepsilon$$ Since$\epsilon$is arbitrary we have a contradiction. 3 deleted 60 characters in body no, it is not true. the following is already contained in Andreas Thom question. from the first paragraph of his question: Let$n$be an integer. Camille Jordan showed that there exists some$m \in {\mathbb N}$(depending on$n$), such that for any pair of$n \times n$-unitaries$u,v \in U(n)$which generate a finite group, one has$[u^m,v^m] = 1_n$. (In fact, he showed that any finite subgroup of$GL(n)$has a abelian normal subgroup of finite index, bounded independently of the subgroup.) Take$u_1,u_2\in{U}(n)$that generate a free group (easy to construct for$d\geq{2}$), and let$m$be as above. Then, since$v_1^m,v_2^m$commute, $$\|u_1^mu_2^m-u_2^mu_1^m\|\leq{}2\|u_1^m-v_1^m\|+2\|u_2^m-v_2^m\|\leq{}4m\varepsilon$$ Taking$\varepsilon$less then the left hand side (not zero by assumption) divided by Since$4m$gives \epsilon$ is arbitrary we have a counterexamplecontradiction.

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