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S3 questionby tyrannosaurus
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#1
May2310, 05:33 PM

P: 37

1. The problem statement, all variables and given/known data
Let G=<a, ba^2=b^2=e, aba=bab>. SHow G is isomorphic to S3. 2. Relevant equations [b]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? 


#2
May2410, 12:41 PM

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P: 4,500

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) 


#3
May2410, 01:37 PM

P: 37

thanks for your help, that helped a lot.



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