I think he showed the example for s3 if you when he calculated aH*bH it had four elements but abH on my had three elements so they can’t be the same. The reason is because once you take the power product of the two sets above. Then in general ahbh is not necessary equal to some abh but when the are normal you can find another h where abh=ahbh FYI those small h are different. Thee reason is aH=Ha. So ah1=h2a so you can commute a up to H I.e it might not be the same h but another h. So product of the two sets will have same size as H
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Concepts Crystal Clear ( CCC rated :) )
Good job
What will be the problem if we take subgroup not necessarily normal and directly define (aH)(bH)=(abH) with considering product of sets?
Sometime (aH)(bH) not belong to left coset of H ie. Cannot be defined as cH if H is not normal. So we cannot take it in general.
Binary operation is itself is not well defined
I think he showed the example for s3 if you when he calculated aH*bH it had four elements but abH on my had three elements so they can’t be the same. The reason is because once you take the power product of the two sets above. Then in general ahbh is not necessary equal to some abh but when the are normal you can find another h where abh=ahbh FYI those small h are different. Thee reason is aH=Ha. So ah1=h2a so you can commute a up to H I.e it might not be the same h but another h. So product of the two sets will have same size as H
Thank you sir
Thanku sir