One important point: We do not need to prove that H intersection N is a normal subgroup of the group homomorphism because we we finally get it as the kernel of the homomorphism map.
@06:20 I think the steps of derivation can be as below (noting that S' = e^(i theta) for theta in [0,2pi]) C*/S' = {rS': r \in R+} = {|z| S': z \in C*} = {|z| e^(i arg(z)) S': z \in C*} = {z S': z \in C*}
Sir, you said about a particular case that G can be thought of as a subgroup of G' when G is homomorphic to G' and Ker (phi) is e only. But sir, how is it possible that if G and G' are defined with different operations..?
Such a great detailed informative video. I can't be more grateful. Best wishes and regards
One important point: We do not need to prove that H intersection N is a normal subgroup of the group homomorphism because we we finally get it as the kernel of the homomorphism map.
@06:20 I think the steps of derivation can be as below (noting that S' = e^(i theta) for theta in [0,2pi])
C*/S'
= {rS': r \in R+}
= {|z| S': z \in C*}
= {|z| e^(i arg(z)) S': z \in C*}
= {z S': z \in C*}
Second Isomorphisam looks like corollary rather than Thereom
Tq sir .. very helpful..
Sir, you said about a particular case that G can be thought of as a subgroup of G' when G is homomorphic to G' and Ker (phi) is e only. But sir, how is it possible that if G and G' are defined with different operations..?
it simply means that the elements of G behave in the same way as the elements of G' do.
@@AvinashKumar-jc3ng Ok.. Thank you..
Is H/N and HN/N same?
H/N is not defined as N is not a subgroup of H
Wow!!!!!
I don't understand that why are there so less views?
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