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The question stands:

Let $S=\mathbb{C}-\{1,0\}$. Describe the subgroup of $\operatorname{Perm}(S)$ generated by the functions: $f:S\rightarrow S, z\mapsto 1-z$ and $g:S\rightarrow S, z\mapsto 1/z$.


I'm having trouble gathering such permutations under $S$ which consist of the complex numbers. Any assistance would be great !

Thank you in advanced.

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Hints:

Well, to begin with, both $\,f,g\,$ are involutions:

$$f^2(z):=f(1-z):=1-(1-z)=z\;,\;\;g^2(z):=g\left(\frac{1}{z}\right):=\frac{1}{\frac{1}{z}}=z$$

Next:

$$fg(z):=f\left(\frac{1}{z}\right):=1-\frac{1}{z}\;,\;\;gf(z):=g(1-z):=\frac{1}{1-z}\Longrightarrow$$

$$gfg(z)=g\left(1-\frac{1}{z}\right):=\frac{1}{1-\frac{1}{z}}=\frac{z}{z-1}$$

$$fgf(z)=f\left(\frac{1}{1-z}\right)=1-\frac{1}{1-z}=\frac{z}{z-1}\ldots$$

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  • $\begingroup$ Thank you so much ! This is great ! $\endgroup$ – Gustavo Louis G. Montańo Mar 26 '13 at 0:50
  • $\begingroup$ Fue un gusto para mí. $\endgroup$ – DonAntonio Mar 26 '13 at 0:57
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\begin{align} \operatorname{id}(z) & = z \\[10pt] f(z) & = 1-z \\[10pt] g(z) & = \frac1z \\[10pt] f(g(z)) & = \frac{z-1}{z} \\[10pt] g(f(z)) & = \frac{1}{1-z} \\[10pt] f(g(f(z))) & = \frac{z}{1-z} \\[10pt] \end{align}

Now go through the possible compositions of these six functions and observe that you never get any more functions besides thes six, and also that every function on this list has an inverse function that is also on this list.

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  • $\begingroup$ Since the set is finite and contains the identity function it is enough to prove that it is closed under composition. $\endgroup$ – lhf Mar 26 '13 at 0:06
  • $\begingroup$ Thank you so much! This is excellent. I worried too much on the complex number itself to look for a neat result in compositions ! $\endgroup$ – Gustavo Louis G. Montańo Mar 26 '13 at 0:49
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The elements of our group is words in the letters $f$ and $g$. For starters, notice that $g(g(z))=z$ and $f(f(z))=z$ so we can ignore pairs of the same letter. Our elements are now alternating strings of $f$ and $g$. The identity is the empty string. Now do they commute? $f(g(z))=1-\frac 1z, g(f(z))=\frac 1{1-z}$, no but $g(f(g(z)))=\frac z{z-1}$ which gives us $f(g(f(g(z))))=1-\frac z{z-1}=\frac 1{1-z}=g(f(z))$. This gives us that $f(g(f(z)))=g(f(g(z)))$ and any word can be reduced to no more than three applications of a function. So our group consists of $\{\emptyset,f,g,fg,gf,fgf\}$ with concatenation the operation and $f^2=g^2=\emptyset, fgf=gfg$

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@eXtremity From the previous answers you must be able to deduce that your subgroup is isomorphic to S$_3$, the symmetric group on three letters.

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