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could any one just tell me can I identify the group $GL_n(\mathbb{R})/H$ with the group $(\mathbb{R}^{+},.)$? where $H=$ Normal subgroup of matrices with positive determinant. .Any correct answers and hint will be appreciated, what I did was just a gues, I dont know how what will be the identification and how.Thank you

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No you cannot. Note that the quotient cannot have more than two elements. –  Ehsan M. Kermani Jan 19 '13 at 16:42
    
I think that is the right guess if $H$ is the group of matrices with $\pm 1$ determinant.... –  N. S. Jan 19 '13 at 17:05

1 Answer 1

up vote 1 down vote accepted

If $n$ is even, then for every $M \in GL_n(\mathbb{R})$, either $M \in H$ or $M=-M'$ with $M'=-M \in H$ (because $\det(-M)=(-1)^n\det(M)=- \det(M)$). So $GL_n(\mathbb{R})/H \simeq \mathbb{Z}_2$.

However, $M \mapsto |\det(M)|$ is an epimorphism from $GL_n(\mathbb{R})$ to $\mathbb{R}_+$ and its kernel is $SL_n(\mathbb{R})= \{ M \in GL_n(\mathbb{R}) : \det(M)=\pm 1\}$, so $GL_n(\mathbb{R})/SL_n(\mathbb{R}) \simeq \mathbb{R}_+$.

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if i use the homomorphism $f(A) = \frac{\det(A)}{|det(A)|}$ where $f :GL_n(R) \rightarrow \{1,-1\}$(both groups under the operation $"."$) and conclude what you have concluded in your first set of arguments about $GL_n(\mathbb{R})/H \simeq \mathbb{Z}_2$ –  jim Jan 21 '13 at 7:29
    
@jim: You should post your comment as an answer. –  Seirios Jan 21 '13 at 8:34
    
@jim thank uuuuuuuuuuuuuuuuuu –  El Angel Exterminador Jun 3 '13 at 17:18

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