I'm wondering how to prove the associativity and identity to prove that the Möbius transformations forms a group.

A Möbius transformation is a complex function of the form $M(z)=\dfrac{az+b}{cz+d}$.

Thank you in advance :)

  • 1
    $\begingroup$ Note that there is the additional requirement that $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ be invertible. Also, the Möbius transformations are a subset of the bijections of the extended complex plane, which definitely are a group. So you only need to show that Möbius transformations are closed under composition and inversion, because then the Möbius transformations are a subgroup. $\endgroup$ Dec 4 '20 at 22:20
  • $\begingroup$ yes! agreed and tysm but would you know how to prove it is a group using associativity and identity $\endgroup$ Dec 4 '20 at 22:22
  • $\begingroup$ Have you tried just multiplying a few of them together? I'm sure you'll quickly see how to show the composition of two is still a Mobius transformation for example. What have you tried? $\endgroup$
    – Dan Rust
    Dec 4 '20 at 22:24
  • $\begingroup$ You don't need to show that they are associative and have an identity. Since they are a subset of a group, they are automatically associative, and if the two closure conditions I mentioned hold, then all other group axioms follow. $\endgroup$ Dec 4 '20 at 22:25

Since you are explicitly asking for the direct proof:

  1. concerning the associativity, this is just the same proof as for the associativity of any group of functions. So consider three moebius transforms $M_1,M_2,M_3$ then for any complex number $z$ we have $$M_1\circ( M_2\circ M_3)(z)=M_1((M_2\circ M_3)(z))=M_1(M_2(M_3(z)))=(M_1\circ M_2)(M_3(z))=(M_1\circ M_2)\circ M_3(z)$$ Note that we did not use any of the properties of the moebius transforms since the associativity just follows from the definition of the concatenation of functions.
  2. For the identity consider the element $M(z)=id(z)=z$, i.e. choose $a=d=1,b=c=0$ then clearly $ad-bc\neq 0$ so this is a moebius transform and you can easily verify that this is the identity for your group.
  3. I guess you already figured out the formula for the inverse transformation since you were only asking for the other two properties, but still for completeness: it is easy to check that for any moebius transform $M(z)=\frac{az+b}{cz+d}$ we have an inverse given by $M^{-1}(z)=\frac{dz-b}{-cz+a}$ and this also is a moebius transform
  4. We actually also have to show that the concatenation of two moebius transforms is again a moebius transform: but $$M_1\circ M_2(z)=M_1(\frac{a_2z+b_2}{c_2z+d_2})=\frac{a_1(\frac{a_2z+b_2}{c_2z+d_2})+b_2}{c_1(\frac{a_2z+b_2}{c_2z+d_2})+d_2}=\frac{(a_1a_2+b_1c_1)z+(a_1b_2+b_1d_2)}{(a_2c_1+c_2d_1)z+(b_2c_1+d_1d_2)}$$ Now denoting the coefficients of the concatenation by $a',b',c',d'$ we have $a'd'-c'b'\neq 0$ which can either be verified by direct calculation or the fact that $0\neq det(\left(\begin{array}{l}a_1&b_1\\c_1&d_1\end{array}\right)\left(\begin{array}{l}a_2&b_2\\c_2&d_2\end{array}\right))=\det(\left(\begin{array}{l}a'&b'\\c'&d'\end{array}\right))$so this claim also holds.

So the set of moebius transforms forms a group under concatenation.
Note that it actually suffices to prove 3 and 4 as this shows that the moebius transforms form a subgroup of the group of all bijections of the extended complex numbers to themselves. But since you were explicitly asking for the first two steps i provided them too :)
Lg Mo

  • 1
    $\begingroup$ You don't even need the fact that the identity is a Möbius transformation (even though it is trivial). If a non-empty subset of a group is closed under concatenation and inversion, then the identity is automatically included since if $b$ is in the set, then so is $b^{-1}$ and therefore $bb^{-1}=1$ as well. $\endgroup$ Dec 4 '20 at 22:27
  • $\begingroup$ I know @Vercassivelaunos, but he explicitly asked for an neutral element etc.(Edited the post to give a complete answer.) $\endgroup$
    – Mo145
    Dec 4 '20 at 22:46
  • $\begingroup$ Thanks for pointing it out, fixed this now too! $\endgroup$
    – Mo145
    Dec 4 '20 at 23:08

I don't know how much you've done on your own, but to help subsequent users, I'll put the full answer. Stop after each hint and try and do it yourself.

Hint 1: Prove that the Möbius transformations are bijective from the extended complex numbers to themselves.

Hint 2: The Möbius transformations are therefore a subset of the set of bijections on the extended complex numbers.

Hint 3: The set of all bijections of the complex numbers is a group.

Hint 4: Composition of two Möbius transformations and inversion of one Möbius transformation is another Möbius transformation.

Hint 5: A closed subset of a group is a group.

Hint 6: The set of Möbius transformations is closed.

  • 1
    $\begingroup$ Note that Möbius transformations are bijections from the extended complex number to themselves, not just the complex numbers themselves. Also, what kind of homomorphism do you mean? They are certainly no field homomorphisms, nor group homomorphisms or vector space homomorphisms. They are conformal, though, and conformal maps from a domain to itself are also called automorphisms. That's a lot more to show than needed, though. Also, to be a group, it must also be closed under inversion, not just composition. $\endgroup$ Dec 4 '20 at 22:43
  • $\begingroup$ I meant the latter, that they were conformal maps from the Riemann sphere to itself. $\endgroup$ Dec 4 '20 at 23:05

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