1. The problem statement, all variables and given/known data Let G=<a, b|a^2=b^2=e, aba=bab>. SHow G is isomorphic to S3. 2. Relevant equations 3. The attempt at a solution[/ since a^2=b^2=e, then |a|=1 or 2 and |b|= 1 or 2. But since aba=bab, the orders of a and b both have to be 2 because if either had order 1, we would get that a=a^2 or that b=b^2. Since aba= bab, by left hand multiplication and right hand multiplication by ab and (ab)^-1 we get that (ab)(aba)(ab)^-1 = abab(ab)^-1= ab. So then ab and aba are conjugates, from this we can partion G by <aba>. So G= a<aba> union b<aba> union e<aba>. Thus G has at most 6 elements (since |aba|=2). SO |G|<= 6. From this, since |S|>=|G| we can show that S3 satisfies the defining relations of G. Is my thinking on this right?
If both a and b are the identity that's not a problem. Identifying rigorously groups represented in this form is kind of tricky, because it's not immediately obvious that everything isn't the identity. The definition is usually something like 'the largest group satisfying these properties' or maybe something involving a lifting property. You should check exactly what your definition is so you can use the definition in your proof This is on the right track. A cleaner proof might start by identifying what a and b are supposed to be in S_{3} then defining a homomorphism from G to S_{3} that takes a and b to their respective elements. A couple of things that you should note that hold true in general (and you should think about if it's not obvious why it's true) 0) The homomorphism must be well defined. In this case it suffices to show that the images of a and b satisfy the properties that a and b satisfy in G 1) If G is in fact S_{3}, the elements that a and b are sent to in S_{3} have to generate S_{3}. Once you prove this you have that your homomorphism is onto. 2) Once 1 is done, all you have to do is check what the kernel is. An argument involving the sizes of the groups can work here (If the map is onto and G has only 6 elements, it has to be injective also)