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How could I show that the 3 groups

$x\to x+a$ $x\to \exp(b) x$ $x\to {x\over 1-cx}$ where $a,b,c\in \mathbb R$ and also $x\in \mathbb R$,

generate a group $x\to {\alpha x+\beta\over \gamma x+\delta}$ where $\alpha\delta-\beta\gamma=1$?

Thank you.


I can see how the 3 smaller groups are subgroups of the bigger group, but I don't know how to show that they generate the larger group.

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    In this case, each of the smaller groups is a subgroup of the larger group. So necessarily, when you compose elements from each of them, you will end up with something in the larger group. (If the three smaller groups weren't subgroups of some larger group, then you wouldn't be able to compose them.)2012-11-26

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Given,

$ f : x \to x+a, \quad g : x \to e^bx, \quad h : x \to \frac{x}{1-cx}, $

then we compose as

$ h(g(f(x)))=h(g(x+a))=h(e^b(x+a))=\frac{e^b(x+a)}{1-c(e^b(x+a))}\,. $

Now, simplify the above and you will get $\alpha,\beta,\gamma,$ and $\delta$.