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Let $G$ be the group of all $n\times n$ matrices with real entries that are invertible. The operations of the group are matrix multiplication, matrix inversion, and identity matrix. Let $S$ be the set of $n\times n$ matrices with real entries and with determinant $1$. Let $R^\times$ denote the group of nonzero numbers with the operations of multiplication, multiplicative inverse, and $1$.

Prove that $G/S \cong R^{\times}$.

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  • $\begingroup$ Are you familiar with the determinant, in particular the fact that it is a homomorphism? $\endgroup$ Commented Apr 14, 2013 at 20:15
  • $\begingroup$ @AlexBecker, it would be very difficult for the OP to know what SL is otherwise... $\endgroup$ Commented Feb 1, 2018 at 19:03

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If you know that $\det:G\to \mathbb R^\times$ is a surjective homomorphism, then the exercise is easy. Observe that $\ker\det = S$ (since $1$ is the identity of $\mathbb R^\times$) and so by the first isomorphism theorem, $\mathbb R^\times\cong G/\ker \det = G/S$.

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  • $\begingroup$ do I have to show that det:$G \rightarrow R^{x}$ is a onto homomorphism or can I just assume? $\endgroup$ Commented Apr 14, 2013 at 20:29
  • $\begingroup$ @Gamecocks99 That depends on whether it has been shown/assumed in your class? $\endgroup$ Commented Apr 14, 2013 at 20:30
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    $\begingroup$ $$\forall\,r\in\Bbb R\;,\;\;\det\begin{pmatrix}r&0&0&\ldots&0\\0&1&0&\ldots&0\\\ldots&\ldots&\ldots& \ldots&\ldots\\0&0&\ldots&0&1\end{pmatrix}=r$$ $\endgroup$
    – DonAntonio
    Commented Apr 14, 2013 at 21:02

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