Why does $PSL(2,\mathbb C)\cong PGL(2,\mathbb C)$ but $PSL(2,\mathbb R) \not\cong PGL(2,\mathbb R)$? Why does $PSL(2,\mathbb C)\cong PGL(2,\mathbb C)$ but $PSL(2,\mathbb R) \not\cong PGL(2,\mathbb R)$?

One way of proving the non-isomorphism part would be to show that ${\rm PGL}(2,{\mathbb R})$ has the Klein 4-group $C_2 \times C_2$ as subgroup, but ${\rm PSL}(2,{\mathbb R})$ does not.
The first claim is easy. $C_2 \times C_2$ is the image in ${\rm PGL}(2,{\mathbb R})$  of the dihedral group of order 8 generated by $\left(\begin{array}{rr}1&0\\0&-1\end{array}\right)$ and $\left(\begin{array}{rr}0&1\\1&0\end{array}\right)$.
It is also straightforward to show that the only element of order 2 in ${\rm SL}(2,{\mathbb R})$ is $-I_2$. So the only possibility for the inverse image in ${\rm SL}(2,{\mathbb R})$ of $C_2 \times C_2$ is the quaternion group $Q_8$. But $Q_8$ does not have a 2-dimensional real representation. That can be shown using the Frobenius-Schur indicator of the 2-dimensional complex representation, but I expect you would prefer a more elementary proof.
What is the source of this problem? Is it an exercise, and if so at what level?
Edit: In fact it is not hard to show that $Q_8$ is not a subgroup of ${\rm GL}(2,{\mathbb R})$ without using representation theory. An element of order 4 in ${\rm GL}(2,{\mathbb R})$ has minimal polynomial $x^2+1$ and is therefore conjugate to $A := \left(\begin{array}{rr}0&1\\-1&0\end{array}\right)$. By simple linear algebra we find that the matrices conjugating $A$ to $A^{-1}$ have the form $B:=\left(\begin{array}{rr}a&b\\b&-a\end{array}\right)$, and $B^2 = -I$ gives $a^2+b^2 = -1$, which has no solution in ${\mathbb R}$. 
So you now have three proofs of the non-isomorphism!
