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 $SL(2,\mathbb{C})$.

  • $\begingroup$ 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. $\endgroup$
    – k.stm
    Mar 21 '14 at 20:40
  • $\begingroup$ 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 $SL(2,\mathbb{C})$ $\endgroup$ Mar 22 '14 at 0:40

Your statement:

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.