# Thread: group of Möbius transformations isomorphic to S3 group

1. ## group of Möbius transformations isomorphic to S3 group

I have problem with understanding the following problem. Show that the group of Möbius transformations is isomorphic to $S_{3}$
We are also supposed to identify involutions i.e elements of order 2.
The hint to the problem is :show that the group permutes the elements $0,1,\infty$ What isomorphism can be constructed here?

2. ## Re: group of Möbius transformations isomorphic to S3 group

Which Mobius transformations, explicitly, are you talking about? The general set of Mobius functions is infinite, you must have a particular 6 in mind....I suspect you mean:

f(x) = x
g(x) = 1/x (this is an involution)
h(x) = 1 - x (this is also clearly an involution)
k(x) = x/(x - 1) (this is not so obviously an involution)
m(x) = 1/(1 - x)
n(x) = (x - 1)/x

to construct an isomorphism with S3, follow the hint: do the involutions fix one of {0,1,∞}? Which functions correspond to 3-cycles?

3. ## Re: group of Möbius transformations isomorphic to S3 group

Originally Posted by Deveno

to construct an isomorphism with S3, follow the hint: do the involutions fix one of {0,1,∞}? Which functions correspond to 3-cycles?
the id function $f(x)=x$ fixes all these 3 elements $0,1,\infty$

$g(x)=\frac{1}{x}$ fixes 1, $\infty$ ?

I still do not know how to construct a function that will be bijective and will give me my isomorphism..

4. ## Re: group of Möbius transformations isomorphic to S3 group

No, g(x) = 1/x does NOT fix ∞, it "swaps" 0 and ∞. If we ORDER our 3-element set like so:

{0,1,∞}

then it stands to reason that g acts on this set the same way the transposition (1 3) acts on the (ordered) set {1,2,3}.

Clearly the identity is f. So if we call our bijective mapping φ: we should have:

φ(f) = id
φ(g) = (1 3)

Any thoughts on how to continue?