Let
$GL_{n}(\mathbb{R}) = \{ A \in M_{n}(\mathbb{R}) \ | \ \text{det}(A) \neq 0 \}$
$SL_{n}(\mathbb{R}) = \{ A \in M_{n}(\mathbb{R}) \ | \ \text{det}(A) = 1 \}$
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 $GL_{n}(\mathbb{R})/H \cong \mathbb{R}$?