Proving G=1: Exercise from Serre's Book Trees

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This is an exercise from a book from Serre called Trees. Given the group

G = < a, b, c | bab−1 = a2, cbc−1 = b2, aca−1 = c2 >

I have to prove G = 1.

I don't have a clue. Of course G' = G (commutator subgroup equals the group itself) but I don't know what to deduce from that. Another first step could be to prove that the orders of a, b and c are finite. But I do not even know how to that.

If anyone could put me in the right direction i would be very grateful.

edit: I am going to try to use Todd Coxeter coset enumeration.
 
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I doubt wetter ToddCoxeter CE will help me :frown:
 
Since the question uses a,b,c and it's Serre, let's assume that the fact there are 3 generators is important.

The only thing I can think of doing is considering products like abc or bac etc and simplifying in two ways until ended up with something that points towards the statement a=a^2, or similar.

(No, I\ve not solved this - I'll get a pen and paper and think about it some more later)
 
well, if you can prove all the 3 generators are of finite order than I can finish this question. Because Suzuki has this exercise in his book where he assumes G is finite and he also gives lots of hints.

Here however I can not a priori assume G is finite.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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