# Isomorphism between Möbius transformations and $SL(2,\mathbb{C})/\mathbb{Z}_2$

This is a problem from A Course in Modern Mathematical Physics by Peter Szekeres.

Here's the quote to the problem I'm solving:

Show that the map $\mu$ from $SL(2,\mathbb{C})$ to the Möbius group is a homomorphism, and that the kernel of this homomorphism is $\{I;-I\}$; i.e the Möbius group is isomorphic to $SL(2,\mathbb{C})/\mathbb{Z}_2$.

What I did was to define

$$\mu:\left( \begin{array}{ccc} a & b \\ c & d \end{array} \right)\rightarrow m(z)=\frac{az+b}{cz+d}$$ where $ad-bc=1$. By taking the image of two unimodular matrix product gives us the composition of two Möbius transformations which is exactly the result we expect, it is easy to see why $\{I;-I\}$ (identity matrix and the "negative" identity matrix) is the kernel of this homomorphism, by the theorem that goes:

If $\varphi: G\rightarrow G'$ is a homomorphism then the factor group $G/\ker(\varphi)$ is isomorphic with the image subgroup $im(\varphi)\subseteq G'$

I can prove that the group of the Möbius transformations is isomorphic to $SL(2,\mathbb{C})/\{I;-I\}$, can't find a way to prove that it is isomorphic to $SL(2,\mathbb{C})/\mathbb{Z}_2$ because i thought that you could only factor a group by a normal subgroup of that group, I don't know if it is possible to factor it by a group wich is isomorphic the normal subgroup of that group. If that were the case, I could prove that $\{I;-I\}$ is isomorphic to $\mathbb{Z}_2$ and then prove that $SL(2,\mathbb{C})/\mathbb{Z}_2 \cong SL(2,\mathbb{C})/\{I;-I\}$ which by transitivity should be isomorphic to the group os Möbius transformations. If this is not posible, i don't see how $\mathbb{Z}_2$ is a normal subgroup of $SL(2,\mathbb{C})$.

• What is your notion of subgroup? Are you thinking about a subgroup in terms of sets? If so, you need to interpret $Z_2$ as a subset of $SL(2,ℂ)$ – what is your interpretation? I think, the question is merely stated a bit clumsy. The question is most likely just about proving $SL(2,ℂ)/\{±I\} \cong \text{Möbius Transformations}$ and you are already done. – k.stm Mar 21 '14 at 20:40
• Yes i see a subgroup as a set where operation between elements is closed and contains an identity element and each element has its inverse, in that sense $\{I;-I\}$ is a subgroup of $SL(2,\mathbb{C})$, I see the elements of $\mathbb{Z}_2$ as scalars, I don't know if they can be interpreted as matrices so I don't see why $\mathbb{Z}_2$ is a subset of $SL(2,\mathbb{C})$ – Mark A. Ruiz Mar 22 '14 at 0:40

If $\varphi: G\rightarrow G'$ is a homomorphism then the factor group $G/ker(\varphi)$ is isomorphic with the image subgroup $im(\varphi)\subseteq G'/ker(\varphi)$
is almost right: the last part should be $im(\varphi)\subseteq G'$. Your map $\mu$ is a surjective homomorphism and hence $im(\mu)$=full Möbius group.