Question regarding the first theorem of isomorphism ?
$b)\ $$\frac{GL_n(\mathbb{Z}) }{SL_n(\mathbb{Z})} \cong \ ?$
$d)\ $$\frac{GL_n(\mathbb{R}) }{SL_n(\mathbb{R})} \cong \ ?$
My atttempt :
For option $d$, the map $f: GL_n(\mathbb{R})\to \{-1,1\}$ given by $f(A) = \mathrm{sign}(\det A)$ is a surjective homomorphism with $\ker f = SL_n(\mathbb{R})$, and so by the first isomorphism theorem, $GL_n({\mathbb{R}})/SL_n(\mathbb{R})\cong \{-1,1\}$.
I don't know the other option $ b)$.
Best Answer
For part b), you could see $GL_2(\mathbb{Z})=\{X|\det(X)=\pm1\}$. Thus if you take your homomorphism $\varphi:GL_2(\mathbb{Z})\to\{1,-1\}$ as $\varphi(X)=\det(X)$ (you don't really need the sign function here since its determinant is a sign...). Now, by proving $\varphi$ is an isomorphism with a kernel $SL_2(\mathbb{Z})$, you could see that $$GL_2(\mathbb{Z})/SL_2(\mathbb{Z})\cong\varphi(GL_2(\mathbb{Z}))=\{+1,-1\}$$
For d), you could consider a more general case $GL_n(\mathbb{F})/SL_n(\mathbb{F})$, where $\mathbb{F}$ is an arbitrary field.
Let the homomorphism, $\varphi:GL_n(\mathbb{F})\to\mathbb{F}$, be $\varphi(X)=\det(X)$ in which you replace all operations in $\mathbb{R}$ with corresponding operations in $\mathbb{F}$. You could see that $\varphi$ has a kernel $SL_n(\mathbb{F})$ since $\forall X\in SL_n(\mathbb{F})$, $\det(X)=1_F$ according to the definition of special linear groups.
Now by the first isomorphism theorem, it follows that $$GL_n(\mathbb{F})/\ker\varphi\cong\varphi(GL_n(\mathbb{F}))$$ Because $\varphi(X)$ can take any value except $0$, we could see $\varphi(\mathbb{F})=\mathbb{F}^\times$ which means $$GL_n(\mathbb{F})/SL_n(\mathbb{F})\cong\mathbb{F}^{\times}$$ or in part d) $$GL_2(\mathbb{R})/SL_n(\mathbb{R})\cong\mathbb{R}\backslash\{0\}$$
Note that $\mathbb{Z}^\times=\{+1,-1\}$, though maybe this notation is not valid since $\mathbb{Z}$ is not a field.