$GL(2,R) / SL(2,R)$ isomorphic to R*

Question

I am needing to write a prove showing that $GL(2,\mathbb{R}) / SL(2,\mathbb{R}) $ is isomorphic to $\mathbb{R}^*$.

I know that $SL(2,\mathbb{R})$ is a normal subgroup of $GL(2,\mathbb{R})$ but I'm not sure how to use that or where I should start.

Any help would be appreciated. Thanks.

Answer 1

The homomorphism $GL(2,\Bbb R)\to\Bbb R^*$, $A\mapsto \det A$ has $SL(2,\Bbb R)$ as kernel.

Answer 2

This works equally well for any $n$, not just $n=2.$ Let $G$ denote non-singular matrices with real entries of size $n\times n$. And let $H$ denote the subset there consisting matrices of determinant 1. You already know $H$ is a normal subgroup.

We can show that the quotient group $G/H$ is isomorphic to $\mathbf{R}^*$. For that we use the fact that $\det(AB)= \det(A)\times \det(B).$

(now I'll start using small letters to denote matrices!) For any coset $gH$, pick an arbitrary representative $x=gh$ where $h\in H$. Define $\phi(gH) = \det (g)$. This is well-defined (ie independent of $h\in H$.)

Can easily check this is a homomorphism of groups.

To show surjectivity of $\phi$: given a non-zero number like $5$ consider the diagonal matrix $y$ with one diagonal entry $5$ and all other diagonal entries $1$. Then $\phi(yH)=5$. Injectivity you can attempt to prove as an exercise.

Answer 3

I present a much simpler proof.

Define $f:GL(2,\mathbf R)\rightarrow \mathbf R^*$ as $A\mapsto \det A.$

Since $SL(2, \mathbf R)$ is a normal subgroup of $GL(2, \mathbf R)$, by first isomorphism theorem, $GL(2,\mathbf R)/SL(2,\mathbf R)\cong f(GL(2,\mathbf R))$,

For any element of $\lambda \in \mathbf R^*$, we have a preimage $A=\begin{bmatrix} \lambda&0\\0&1\end{bmatrix} \in GL(2,\mathbf R)$ of function $f$.

Derek Holt on 12th October kindly pointed out this simple fact: $f\left (\begin{bmatrix} \lambda&0\\0&1\end{bmatrix}\right) = \lambda$, it implies $f(GL(2,\mathbf R)) = \mathbf R^*$.

Thus, $GL(2,\mathbf R)/SL(2,\mathbf R)\cong \mathbf R^*$, from first isomorphism theorem.