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I am asking purely out of interest: What the abelianization of general linear group $GL(n,\mathbb{R})$?

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The derived group of ${\rm GL}(n,K)$ is ${\rm SL}(n,K)$ for all $n \ge 1$ and all fields $K$, so the abelianization of ${\rm GL}(n,K)$ is the multiplicative group of the field, $(K \setminus \{0\},\times)$.

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    $\begingroup$ Except when $n=2$ and $K = \mathbb{F}_2$. In that case, $GL(2,\mathbb{F}_2)$ is isomorphic to $\Sigma_3$ and its abelianization in $\mathbb{Z}/2$. $\endgroup$ – Goa'uld Jun 10 '16 at 23:54
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    $\begingroup$ I know this is an old post - but it's interesting to point out that except in the exceptional case, the determinant morphism $GL(n, K) \to K^*$ is a universal abelianization morphism. $\endgroup$ – Daniel Schepler May 11 '17 at 22:46
  • $\begingroup$ Thank you for your answer! $\endgroup$ – Zuriel Apr 7 '18 at 17:36

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