Let G=<a, b|a^2=b^2=e, aba=bab>. To what familiar group is G isomorphic
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, and |b|=2, we get that b^-1(aba)b=a. So then a and aba are conjugates, from this we can partion G. 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?