sir, what you said @ 05:48 in this lecture contradicts from lecture 06 (13:15 - 14:20). if 1 cannot generate 0, the identity element then how can be a group ?
I'm not sure if my doubt is the same as urs, @ 05:48 he include 0 in the sub group generated by , how is that possible? Subgroup generated by 'a' can only contain a^n, where n belongs to Z. When n=0, 1^0 is still 1 right? How does '0' come in the subgroup generated by then?
Here's what I think, Actually the more general way of defining a cyclic group is G ={...,(a^-1)*(a^-1),(a^-1),e,a,a*a,...} under operation * . Here the operation is + and a=1 and we have {...,-3,-2,-1,0,1,2,3,....}(Remember a^-1 means additive inverse here and not our usual inverse like 2^-1 = ½ !!). If you wonder how the 0 comes, note the definition have the identity element e and for addition e = 0. Also, interestingly it is in agreement with the previous theorem, which states "Every subgroup of Z is of form aZ for some non-negative integer a€Z" and here's a=1.
Thank you for excellent lectures. Where can we find the notes that you have written in the course?
sir, what you said @ 05:48 in this lecture contradicts from lecture 06 (13:15 - 14:20). if 1 cannot generate 0, the identity element then how can be a group ?
I'm not sure if my doubt is the same as urs, @ 05:48 he include 0 in the sub group generated by , how is that possible? Subgroup generated by 'a' can only contain a^n, where n belongs to Z. When n=0, 1^0 is still 1 right? How does '0' come in the subgroup generated by then?
@@allenfdo8813 no, a^0= identity element
Here's what I think,
Actually the more general way of defining a cyclic group is G ={...,(a^-1)*(a^-1),(a^-1),e,a,a*a,...} under operation * . Here the operation is + and a=1 and we have {...,-3,-2,-1,0,1,2,3,....}(Remember a^-1 means additive inverse here and not our usual inverse like 2^-1 = ½ !!). If you wonder how the 0 comes, note the definition have the identity element e and for addition e = 0.
Also, interestingly it is in agreement with the previous theorem, which states "Every subgroup of Z is of form aZ for some non-negative integer a€Z" and here's a=1.
={n*1| n€ Z} so 0 €
Great session
Sincere request to delve into difficult topics like fields and rings.
@@lalkish95 Thanks man. You're a life saver
@@lalkish95 Do you know which text is the professor using here? It's certainly not galian as things are proved in a different manner there.
@@yashj1072 What did Lalith Kishore right? It seems he has deleted his comment
@@thatman3107 Professor has a separate playlist on Ring theory
@@yashj1072 oh. Thanks
Thank you sir
THANK YOU SO MUCH !!!
are you preparing for net