$PSl_n(\mathbb{C})\cong PGl_n(\mathbb{C})$? I was reading about projective linear groups because I was asked to show that $PSl_n(\mathbb{C})\cong PGl_n(\mathbb{C})$. Here $PSl_n(\mathbb{C})$ is the projective space of $Sl_n(\mathbb{C})$, i.e. $Sl_n(\mathbb{C})/(\text{scalar matrices in } Sl_n(\mathbb{C}))$ and $PGl_n(\mathbb{C})$ is the projective space of $Gl_n(\mathbb{C})$, i.e. $Gl_n(\mathbb{C})/(\text{scalar matrices in } Gl_n(\mathbb{C}))$. the inclusion map decends to a map that is well defined and injective $$\frac{Sl_n(\mathbb{C})}{(\text{scalar matrices in } Sl_n(\mathbb{C}))} \to \frac{Gl_n(\mathbb{C})}{(\text{scalar matrices in } Gl_n(\mathbb{C}))}$$. But I am not sure about surjectivity. If I take $A \in Gl_n(\mathbb{C})$, I know there is a scalar matrix $S$ such that $det(AS)=\pm 1$ but $Sl_n(\mathbb{C})$ are matrix with $+1$ determinant.
 A: I am spelling out my comments as an answer.
To your first question, if $\det(AS) = -1$, then replace $S$ by $ST$, where $T = \lambda E_n$ is a matrix such that $\det(T) = \lambda^n = -1$. Such a $\lambda$ exists, because the equation $\lambda^n = -1$ is solvable over the complex field.
As noted in the comments, one could take $S$ such that $\det(S) = 1/\det(A)$ from the beginning.
To the second question, which arose in the comments:
Yes, it is essential that $\mathbb C$ is algebraically closed. In general, if one wants to  prove that $\mathrm{PSL}_n(k)$ and $\mathrm{PGL}_n(k)$ are isomorphic for a fixed $n$, it is sufficient that $n$-th roots of all elements of $k$ exist and are contained in $k$. For example, if $n$ is odd, then $\mathrm{PSL}_n(\mathbb R)$ and $\mathrm{PGL}_n(\mathbb R)$ are isomorphic, because you can take $n$-th roots of any real number and still get a real number.
If however $n$ is even, say $n = 2$, it is no longer true that $\mathrm{PSL}_2(\mathbb R)$ and $\mathrm{PGL}_2(\mathbb R)$ are isomorphic. To see this, a little bit of group theory is needed (maybe someone has a simpler argument). Note that $\mathrm{PGL}_2(\mathbb R)$ contains the group
$$\left\langle \left[\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix} \right], \left[\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} \right]\right\rangle\cong \mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z,$$
so that it suffices to prove that $\mathrm{PSL}_2(\mathbb R)$ does not contain a subgroup isomorphic to $\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$. Indeed, if $C := \begin{pmatrix}a & b \\ c & d\end{pmatrix}\in \mathrm{SL}_2(\mathbb R)$ is an element of order $2$, then
$$C^2 = \begin{pmatrix}a & b \\ c & d\end{pmatrix}^2 = \begin{pmatrix}a^2+bc & b(a+d) \\ c(a+d) & d^2+bc\end{pmatrix} = I_2.$$
Note that we also have $\det(C) =ad-bc = 1$. Hence, if $a = -d$, we would have $-a^2-bc = 1$, hence the diagonal entries of $C^2$ are $-1$. Thus $b = c = 0$, and then $C =-I_2$. This proves that the only element of order $2$ in $\mathrm{SL}_2(\mathbb R)$ is $-I_2$.
Assume now that $\mathrm{PSL}_2(\mathbb R)$ has a subgroup $H$ isomorphic to $\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$. Let
$$q \colon \mathrm{SL}_2(\mathbb R) \to \mathrm{PSL}_2(\mathbb R)$$
be the quotient map.
Write $H' := q^{-1}(H)$. Since $q$ is surjective, $q(H')=H$, and hence $|H'| = 8$ (since $\mathrm{ker}(q) = \langle -I_2\rangle$ has order $2$). Moreover, the only element in $H'$ whose order is $2$ is $-I_2$. By the classification of groups of order $8$, we have that
$$H' \cong Q_8 = \langle a,b \, |\, a^4 = 1, a^2 = b^2, ab = b^{-1}a\rangle,$$
the quaternion group of order $8$.
So what we've done so far, is to show the existence of a subgroup $H' \subset \mathrm{SL}_2(\mathbb R)$ which is isomorphic to $Q_8$. Now we somehow need to show, that this is not possible. I don't know your background in representation theory, but it can be shown that $Q_8$ does not have any $2$-dimensional faithful irreducible real representation: the only $2$-dimensional irreducible representation of $Q_8$ is, up to isomorphism,
$$a \mapsto A := \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}, \;\;\;b \mapsto B := \begin{pmatrix} i & 0\\ 0 & -i\end{pmatrix}.$$
To show that this is not isomorphic to any real representation, there are two possibilities:

*

*Use the Frobenius-Schur indicator

*Show by hand that there is no $S \in \mathrm{GL}_2(\mathbb C)$ such that both $S^{-1}AS$ and $S^{-1}BS$ are real matrices.

We conclude that $\mathrm{SL}_2(\mathbb R)$ does not have a subgroup isomorphic to $Q_8$, which means that $\mathrm{PSL}_2(\mathbb R)$ and $\mathrm{PGL}_2(\mathbb R)$ cannot be isomorphic.
