A question about the quotient group $GL(2,\mathbb{C})/\{\lambda I\}$ My question is that : Is $GL(2,\mathbb{C})/\{\lambda I\}$ isomorphic to $PSL(2,\mathbb{C})$ ? If yes, how can one prove it ?
I was thinking in this direction : Given any arbitrary matrix $A$ in $GL(2,\mathbb{C})$, $det(A)=\eta\neq0$. Now In the quotient group $GL(2,\mathbb{C})/\{\lambda I\}$, the matrices $A$ and $A.(\frac{1}{\sqrt{\eta}}I)$ are the same, hence $\forall A\in GL(2,\mathbb{C})/\{\lambda I\}$, $det(A)=1$ Therefore $$GL(2,\mathbb{C})/\{\lambda I\}\cong PSL(2,\mathbb{C})$$ But this does not seem correct, because I could have said : in $GL(2,\mathbb{C})/\{\lambda I\}$, the matrices $A$ and $A.(\frac{1}{\sqrt{2\eta}}I)$ are the same, hence $\forall A\in GL(2,\mathbb{C})/\{\lambda I\}$, $det(A)=\frac{1}{2}$ 
I am confused here...
I already appreciate your helpful comments/answers about the question itself AND the way I looked at at it.
 A: Let $n\in\mathbf{N}$. We define $PGL(n,\mathbf{C})$ to be the quotient groupe of $GL(n,\mathbf{C})$ by its center $Z_1$. We define $PSL(n,\mathbf{C})$ to be the quotient group of $SL(n,\mathbf{C})$ by its center $Z_2$.
By definition $Z_1$ is the set of elements of $GL(n,\mathbf{C})$ commuting to all matrices of $GL(n,\mathbf{C})$. If two matrices $A$ and $B$ commute, the eigenspaces of $A$ are stable by $B$. Therefore an element of $Z_1$ leaves stable any eigenspace of any element of $GL(n,\mathbf{C})$, a fortiori any line of $\mathbf{C}^2$. It is well-known that an endomorphism leavin every line stable is an homothety, and our endomorphism being invertible means that our center elements is $\lambda Id$ for $\lambda\in\mathbf{C}^{\times}$. Therefore : $Z_1 = \mathbf{C}^{\times} Id \simeq \mathbf{C}^{\times}$.
By definition $Z_2$ is the set of elements of $SL(n,\mathbf{C})$ commuting to all matrices of $SL(n,\mathbf{C})$. Let $g\in Z_2$ and $h\in GL(n,\mathbf{C})$. Let $\zeta$ a $n$-root of $det(g)$ and $h_2 := \frac{1}{\zeta} h$. Then $det(h_2) = 1$, showing that $h_2 \in SL(n,\mathbf{C})$. Then $g$ commutes to $h_2$, which implies that $g$ commutes to $h$ also. Therefore $Z_2 \subseteq Z_1$. This shows that $Z_2 = \mathbf{U}_n Id \simeq \mathbf{U}_n$.
Now consider the obvious inclusion $ i : SL(n,\mathbf{C}) \rightarrow GL(n,\mathbf{C})$ mapping an element $g$ to itself. It is a group morphism. As we have $i(Z_2) \subseteq Z_1$, the morphism $i$ induces a so-called quotient morphism $j : PSL(n,\mathbf{C}) = SL(n,\mathbf{C}) / Z_2 \rightarrow GL(n,\mathbf{C}) / Z_1 = PGL(n,\mathbf{C})$ sending the class of $g \in SL(n,\mathbf{C})$ to the class of $g \in GL(n,\mathbf{C})$. The morphism $j$ is obviously injective.
Now let $\theta$ be an element of $PGL(n,\mathbf{C})$. $\theta$ is the class of an element $g\in GL(n,\mathbf{C})$ modulo $Z_1$. We can write $g = det(g) g_2$ with $g_2 \in SL(n,\mathbf{C})$. Multiply it by $\zeta^{-1} Id \in Z_1$ where $\zeta$ is an $n$-rooth of $1$ and you'll get a matrix $M$ of determinant $1$, whose class modulo $Z_1$ is still $\theta$. Note $\eta$ the class of $M$ modulo $Z_2$. Then, $j(\eta) = \theta$ by construction of $j$.
This shows that the canonical (injective) morphism $j : PSL(n,\mathbf{C}) \rightarrow PGL(n,\mathbf{C})$ is in fact an isomorphism.
Taking $n=2$, you have the answer to your question.
For a general $n$, all remains true is $\mathbf{C}$ is replaced by a commutative field in which each element has an $n$-root.
