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 \(\textrm{GL}(n,\Bbb R)/\textrm{SL}(n,\Bbb R)\simeq \Bbb R^\ast\)

**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 \(G\).

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