GLn(R)/SL_n(R) is Isomorphic to R^x

# GLn(R)/SL_n(R) is Isomorphic to R^x

Recall that the general linear group $\mathrm{GL}_n(\mathbb{R})$ is the group of all $n \times n$ invertible matrices with entries in $\mathbb{R}$ with the operation of matrix multiplication, and the special linear group $\mathrm{SL}_n(\mathbb{R})$ is the group of all $n \times n$ invertible matrices with entries in $\mathbb{R}$ and with determinant $1$.

We will now prove that $\mathbb{SL}_n(\mathbb{R})$ is a normal subgroup of $\mathrm{GL}_n(\mathbb{R})$ and that $\mathrm{GL}_n(\mathbb{R})/\mathbb{SL}_n(\mathbb{R})$ is isomorphic to $(\mathbb{R}^{\times}, \cdot)$.

Proposition 1: For each $n \in \mathbb{N}$, $\mathrm{GL}_n(\mathbb{R})/\mathbb{SL}_n(\mathbb{R})$ is isomorphic to $\mathbb{R}^{\times}$. |

**Proof:**Let $\phi : \mathrm{GL}_n(\mathbb{R}) \to \mathbb{R}^{\times}$ be defined for all $A \in \mathrm{GL}_n(\mathbb{R})$ by:

\begin{align} \quad \phi(A) = \det (A) \end{align}

- Observe that $\phi(A) \in \mathbb{R}^{\times}$ since $A \in \mathrm{GL}_n(\mathbb{R})$ if and only if $\det(A) \neq 0$.

- Now observe that:

\begin{align} \quad \ker(\phi) = \{ A \in \mathrm{GL}_n(\mathbb{R}) : \det(A) = 1 \} = \mathrm{SL}_n(\mathbb{R}) \end{align}

- So $\mathrm{SL}_n(\mathbb{R})$ is the kernel of the homomorphism $\phi : \mathrm{GL}_n(\mathbb{R}) \to \mathbb{R}^{\times}$. Thus $\mathrm{SL}_n(\mathbb{R})$ is a normal subgroup of $\mathrm{GL}_n(\mathbb{R})$.

- By The First Isomorphism Theorem we have that:

\begin{align} \quad \mathrm{GL}_n(\mathbb{R})/\mathrm{SL}_n(\mathbb{R}) = \mathrm{GL}_n(\mathbb{R})/\ker(\phi) \cong \phi(\mathrm{GL}_n(\mathbb{R})) \end{align}

- Observe that $\phi$ is surjective. Indeed, for all $x \in \mathbb{R}^{\times}$, consider the $n \times n$ matrix $A_x$ that is defined as a copy of the $n \times n$ identity matrix, but with the entry in row $1$ column $1$ replaced with $x$. Then $\det(A_x) = x$, so $\phi(A_x) = x$. Thus $\phi(\mathrm{GL}_n(\mathbb{R}) = \mathbb{R}^{\times}$, and from above we see that:

\begin{align} \quad \mathrm{GL}_n(\mathbb{R})/\mathrm{SL}_n(\mathbb{R}) \cong \mathbb{R}^{\times} \quad \blacksquare \end{align}