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Let $SL_n(\mathbb{R})$ be the set of $n\times n$ matrix with determinant one. Where can I find a proof that $[SL_n(\mathbb{R}),SL_n(\mathbb{R})]=SL_n(\mathbb{R})$? I've seen this result mentioned in several places, but couldn't find a proof.

For reference, the commutator $[SL_n(\mathbb{R}),SL_n(\mathbb{R})]$ is defined as the group generated by the set of all elements of the form $a^{-1}b^{-1}ab$, with $a,b\in SL_n(\mathbb{R})$.

Reveillark
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  • Wouldn't elements of the commutator have zero trace? Thus for example, the commutator wouldn't contain the identity matrix. Or am I missing something? – User8128 Jul 26 '17 at 23:24
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    @User8128 I don't think so, the commutator is the subgroup generated by all elements of the form $a^{-1}b^{-1}ab$. – Reveillark Jul 26 '17 at 23:26
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    oh I see. I was under the impression that the commutator was the elements of the form $[A,B] =AB -BA$. I suppose I was viewing $SL_n(\mathbb R)$ as an associative algebra rather than a group (and hence using the wrong bracket). Otherwise it's possible I'm just completely mistaken. – User8128 Jul 26 '17 at 23:32
  • @User8128 I'll add the definition to my question to avoid future misunderstandings. – Reveillark Jul 26 '17 at 23:34

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Note that this makes $SL_n(\mathbb{R})$ what's called a perfect group: https://en.wikipedia.org/wiki/Perfect_group

A proof of this fact can be found at GroupProps: https://groupprops.subwiki.org/wiki/Special_linear_group_is_perfect

smb3
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