Flawed isomorphism?

Is there a flaw in the proof below ?

Claim: $\textrm{GL}(n,\Bbb R)/\textrm{SL}(n,\Bbb R)\simeq \Bbb R^\ast~\forall~n\in\Bbb{Z^+}$

Proof.

Step-1) We define a map $\phi\colon\textrm{GL}(n,\Bbb R)\to\Bbb R^\ast$ by $\phi(x)=|x|$ , the determinant of $x$ .

Step-2) We note that $\phi$ is a group homomorphism since $\phi(xy)=|xy|=|x||y|=\phi(x)\phi(y)~\forall~x,y\in\textrm{GL}(n,\Bbb R)$ .

Step-3) $\forall~x\in\Bbb R^\ast~,~\exists y=x^{1/n}\cdot I_n\in\textrm{GL}(n,\Bbb R)$ such that $\phi(y)=(x^{1/n})^n=x$ , so $\phi$ is also an epimorphism.

Step-4) By definition of kernel, we see that $\textrm{Ker}(\phi)=\textrm{SL}(n,\Bbb R)$ since $|X|=1=e_{\Bbb R^\ast}\iff X\in\textrm{SL}(n,\Bbb R)$ .

Step-5) By the first group isomorphism theorem , we conclude that $\textrm{GL}(n,\Bbb R)/\textrm{SL}(n,\Bbb R)\simeq \Bbb R^\ast$

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Details and Assumptions:

• A flaw may be any part of a sentence that is false, for example, even if the conclusion made in a step is correct but the justification provided is false (not true in general), the step would be marked as flawed.
• For two groups $G$ and $H$ , we write $G\simeq H$ to denote that $G$ is isomorphic to $H$ , i.e., there exists a group isomorphism from $G$ to $H$ .
• $\textrm{GL}(n,\Bbb R)$ is the general linear group of the set of invertible square matrices of order $n$ with real entries under the operation of matrix multiplication.
• $\textrm{SL}(n,\Bbb R)$ is the special linear group of the set of square matrices of order $n$ with real entries having determinant $1$ under the operation of matrix multiplication.
• $\Bbb R^\ast$ is the group of non-zero reals under the operation of multiplication of reals.
• $I_n$ is the identity matrix of order $n$ .
• $e_G$ denotes the identity element of group $G$ .