Identifying quotient groups of matrices 
Let $\newcommand{\GL}{\mathrm{GL}}H_n = \{A \in \GL(n,\Bbb C) : \lvert\det(A)\rvert = 1\}$. Prove that $H_n$ is a normal subgroup of  $\GL(n,\Bbb C)$ and that $\newcommand{SL}{\mathrm{SL}} \SL(n,\Bbb C)$ is a normal subgroup of $H_n$. Identify the quotient groups, $\GL(n,\Bbb C)/H_n$ and $H_n/\SL(n,\Bbb C)$, up to isomorphism.

I was trying to solve the problem above. I've shown that $H_n$ is a normal subgroup of  $\GL(n,\Bbb C)$ and that $\SL(n,\Bbb C)$ is a normal subgroup of $H_n$ but I'm not so familiar with quotient groups so I cannot identify them. I've thought $H_n/\SL(n,\Bbb C)$ as $\{I,-I\}$ where $I$ is identity matrix but I'm really not sure. 
I appreciate if someone explain me these two quotient groups.
 A: The key is to use the first isomorphism theorem.
1) Look at the restriction $\det: H_n \to \Bbb S^1$. Then $\ker \det  = {\rm SL}(n,\Bbb C)$ by definition (and we don't run into issues here since ${\rm SL}(n, \Bbb C)\subseteq H_n$). Moreover, even this restricted $\det$ is surjective, since given $z \in \Bbb S^1$, we may take any $n$-th root of $z$, say $z^{1/n}$ (there is more than one choice) and consider $z^{1/n}{\rm Id}_n \in H_n$, which has $\det (z^{1/n}{\rm Id}_n) = z$. This shows that $H_n/{\rm SL}(n,\Bbb C) = \Bbb S^1$.
2) For ${\rm GL}(n, \Bbb C)/H_n$ there is a more interesting geometric interpretation: look at the stratification $${\rm GL}(n,\Bbb C) = \bigsqcup_{z \in \Bbb C^*} {\det}^{-1}(z).$$Every matrix in ${\rm GL}(n, \Bbb C)$ has determinant determined by a direction and positive radius (polar coordinates, $\Bbb C^* = \Bbb S^1 \times \Bbb R_{>0}$). Multiplying a matrix in ${\rm GL}(n,\Bbb C)$ by an element of $H_n$ moves its determinant along the circle where it lies, without changing the radius. And conversely, two matrices whose determinants lie in the same circle are related by an element of $H_n$. More precisely, given $A,B \in {\rm GL}(n,\Bbb C)$ such that $|\det A| = |\det B|$, then $$|\det(AB^{-1})| = \frac{|\det A|}{|\det B|} = 1$$and $AB^{-1} \in H_n$.  So modding out $H_n$ kills the dependence on the direction (i.e., on the $\Bbb S^1$ factor). This indicates that ${\rm GL}(n,\Bbb C)/H_n \cong \Bbb R_{>0}$. Of course this is formally proven by looking at the homomorphism $|\det|:{\rm GL}(n,\Bbb C)\to \Bbb R_{>0}$, noting that it is surjective with $\ker|\det|=H_n$.
