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Let $GL_{n}(\mathbb{F})$ be the set of all $n \times n$ invertible matrices over a field $\mathbb{F}$ of characteristic $0$.

Whether $GL_{r}(\mathbb{F})$ is isomorphic to $GL_{s}(\mathbb{F})$ for $r \neq s$? I think that $GL_{r}(\mathbb{F}) \not\cong GL_{s}(\mathbb{F})$ for $r \neq s$, but I don't know how to prove it. I need to a strict proof.

bing
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    Related on MO (this is a different question): http://mathoverflow.net/questions/106838/isomorphic-general-linear-groups-implies-isomorphic-fields – Watson Aug 29 '16 at 09:14
  • @ Watson Thank you very much. – bing Aug 29 '16 at 09:16
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    $GL_1(F) = F^{\times}$ is abelian, while $GL_2(F)$ is not. – Watson Aug 30 '16 at 08:17
  • Related: http://math.stackexchange.com/questions/1908104/give-a-simple-proof-gl-1-mathbbf-not-cong-gl-2-mathbbf – Watson Aug 30 '16 at 08:20
  • @ Watson https://math.stackexchange.com/questions/1686695/is-there-a-simple-way-of-proving-that-textgl-nr-not-cong-textgl-mr For $m \neq n$, $m,n \geq 2$, $\text{GL}{m}(\mathbb{F})$ and $\text{GL}{n}(\mathbb{F})$ are not elementarily equivalent, how we get $\text{GL}{m}(\mathbb{F}) \not\cong \text{GL}{n}(\mathbb{F})$? – bing Aug 30 '16 at 08:42
  • Let us continue this discussion in chat. – bing Aug 30 '16 at 08:49

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