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Let $G$ be the group of all $3\times 3$ 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 $3\times 3$ matrices with real entries and with determinant of $1$. 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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Are you familiar with the determinant, in particular the fact that it is a homomorphism? –  Alex Becker Apr 14 '13 at 20:15

1 Answer 1

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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do I have to show that det:$G \rightarrow R^{x}$ is a onto homomorphism or can I just assume? –  Gamecocks99 Apr 14 '13 at 20:29
    
@Gamecocks99 That depends on whether it has been shown/assumed in your class? –  Alex Becker Apr 14 '13 at 20:30
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$$\forall\,r\in\Bbb R\;,\;\;\det\begin{pmatrix}r&0&0&\ldots&0\\0&1&0&\ldots&0\\\ldots&\ldots&\ldots& \ldots&\ldots\\0&0&\ldots&0&1\end{pmatrix}=r$$ –  DonAntonio Apr 14 '13 at 21:02

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