If anyone is wondering since a’=ah1 because there H is normal and since aH=a’H same for b then ahbhH=abh3h2H=abH since you can just put it on H for all the h3 and 4
I think there is no need to check whether the operation is well-defined because we defined it in terms of product of sets and not in terms of the product of representatives.
You are right If anyone is wondering since a’=ah1 because there H is normal and since aH=a’H same for b then ahbhH=abh3h2H=abH since you can just put it on H for all the h3 and h4
Thanks a lot sir, really difficult to understand why there are only 11 k views on this video. This is a masterclass from you
Tysm sir for such a valuable information
Good content
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Gr8 sir
If anyone is wondering since a’=ah1 because there H is normal and since aH=a’H same for b then ahbhH=abh3h2H=abH since you can just put it on H for all the h3 and 4
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God bless you
Thanku sir
Thank you sir
I think there is no need to check whether the operation is well-defined because we defined it in terms of product of sets and not in terms of the product of representatives.
You are right If anyone is wondering since a’=ah1 because there H is normal and since aH=a’H same for b then ahbhH=abh3h2H=abH since you can just put it on H for all the h3 and h4