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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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Are you asking what you are supposed to do? You need to decompose an arbitrary transformation in your larger group as a composition of transformations from one of the three smaller groups. To figure out exactly how to do it, you have to experiment a bit and start by seeing what composing two transformations from two of the smaller groups looks like. –  Michael Joyce Nov 26 '12 at 2:29
    
Thank you, @MichaelJoyce. So to show that the smaller groups generate a larger group I just have to show that any element of the larger group CAN be obtained as a composition of elements from the smaller groups? So the larger group is does not necessarily have to contain all the possible transformations obtainable by composing elements from the smaller groups? –  George Nov 26 '12 at 2:49
    
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.) –  Michael Joyce Nov 26 '12 at 4:39

1 Answer 1

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 $\delta$.

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