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I read somewhere that $PGL_2(\mathbb{C})=SL_2(\mathbb{C})/N$ where $N$ is the normal subgroup consisting of $\pm \left( \begin{matrix} 1 & 0\\ 0 & 1 \end{matrix}\right)$. It is unclear to me how this follows from the fact that $PGL_2(\mathbb{C})=GL_2(\mathbb{C})/aI$ where $aI$ are the scalar matrices ($I$ the identity and $a\in\mathbb{C}^*$). However, I did see somewhere the statement that $SL_2(\mathbb{C})/N$ is called $PSL_2(\mathbb{C})$ and that $PGL_2(\mathbb{C})\cong PSL_2(\mathbb{C})$.

Two questions:

1) How does one show that $PGL_2(\mathbb{C})\cong PSL_2(\mathbb{C})$?

2) Is there another (perhaps easier) way of seeing that $PGL_2(\mathbb{C})=SL_2(\mathbb{C})/N$?


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up vote 3 down vote accepted

Hint: If $M \in GL_2(\mathbb{C})$, let $\alpha$ be a square root of $\det(M)$. Then $\frac{1}{\alpha}M \in SL_2(\mathbb{C})$.

Notice that the equality is false over $\mathbb{R}$.

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Ah, of course -- thanks! – Nilay Kumar Jun 12 '13 at 3:27

Hints: Given an element $ M \in {\rm GL}(2, \mathbb{C})$, you can find a complex number $\lambda$ so that $\lambda M$ has determinant $1$. Note that the image of $M$ is the same as the image of $\lambda M$ in ${\rm PGL}(2,\mathbb{C}).$

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Got it, thanks! – Nilay Kumar Jun 12 '13 at 3:27

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