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  • $GL_{n}(\mathbb{R}) = \{ A \in M_{n}(\mathbb{R}) \ | \ \text{det}(A) \neq 0 \}$

  • $SL_{n}(\mathbb{R}) = \{ A \in M_{n}(\mathbb{R}) \ | \ \text{det}(A) = 1 \}$

We know that $$GL_{n}(\mathbb{R})/SL_{n}(\mathbb{R}) \cong \mathbb{R}^{\ast}$$

I am interested to know the answer for this question.

  • Does there exist a subgroup $H$ such that $GL_{n}(\mathbb{R})/H \cong \mathbb{R}$?
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up vote 11 down vote accepted

$\mathbb{R}^{\ast} \simeq \{ \pm 1 \} \times \mathbb{R}$: the isomorphism is gotten by looking at sign and taking the logarithm of the absolute value. So $\mathbb{R}^{\ast}$ has a quotient isomorphic to $\mathbb{R}$, and the kernel of the corresponding composition (the group of matrices with determinant $\pm 1$) provides the desired $H$.

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You think very fast. – anonymous Jan 22 '11 at 20:43
do you mean logarithm of the absolute value? – Dylan Wilson Jan 23 '11 at 0:39
@Dylan: yes, sorry, that's what I meant. – Qiaochu Yuan Jan 23 '11 at 0:45

Well, another way of thinking about this.

  • Consider the mapping from $\text{GL}_{n}(\mathbb{R}) \to (\mathbb{R}_{+},\times)$, $A \mapsto (\text{det}\:(A))^{2}$ and note that $(\mathbb{R},+) \cong (\mathbb{R}_{+},\times)$.

  • Consider the mapping from $\text{GL}_{n}(\mathbb{R}) \to (\mathbb{R}_{+},\times)$, $A \mapsto |\text{det}\:(A)|$ and note that $(\mathbb{R},+) \cong (\mathbb{R}_{+},\times)$.

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