Let $GL(n,\mathbb{R})$ be the group of invertible $n \times n$ real matrices, let $SL(n,\mathbb{R})$ be the group of $n \times n$ real matrices of determinant $1$, and $\mathbb{R}^*$ be the group of nonzero real numbers under multiplication.
Prove: $GL(n,\mathbb{R})/SL(n,\mathbb{R})$ is isomorphic to $\mathbb{R}^*$.
Attempt: We can define a surjective homomorphism
$f: GL(n,R) \to \mathbb{R}^*$. Let $f(A) = \det(A)$. Then
$$
f(AB) = \det(AB) = \det(A)\det(B) = f(A)f(B)
$$
so $f$ is a group homomorphism. And it is surjective since any $r \in \mathbb{R}^*$ is in the image of $f$ since $1/\det A$ is in $\mathbb{R}^*$. We need to show $SL(n,\mathbb{R})$ is the kernel.
Now I already proved in another exercise, $SL(n,\mathbb{R})$ is the kernel, which is a normal subgroup.Then by the first isomorphism theorem, there is an isomorphism
$g: GL(n,\mathbb{R})/SL(n,\mathbb{R})→ \mathbb{R}^*$. Thus, we can conclude $GL(n,\mathbb{R})/SL(n,\mathbb{R})$ is isomorphic to $\mathbb{R}^*$.
Can anyone please check my proof? Any help/feedback would be really appreciated. Thank you.
Best Answer
What you've done is show the image is closed under inverses. If a set-theoretic map $f:G\to H$ satisfies $f(ab)=f(a)f(b)$ for all $a,b\in G$ then it is automatically a group homomorphism and the image is closed under inverses - this is a priori generically true for all groups and homomorphisms and is irrelevant to arguing surjectivity and irrelevant to this exercise.
You need to prove surjectivity. This means given a nonzero real number $r$ you need to prove there exists a matrix $A\in M_{n\times n}(\Bbb R)$ such that $\det A=r$. As a hint, handle the cases $r>0$ and $r<0$ separately according to whether $n=1$ or $n\ge2$.