# Flawed isomorphism?

Algebra Level 4

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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