Thank you for explaining why r=0! My textbook didn't explain that part. Also, I like that we can see you in the video. I don't know why, but it helped me understand what you were saying.
awesome , well explained. What if you started with G = , which implies that G = {e} since e^n = e for any integer n. Then it is trivial to show that any subgroup of G is cyclic, since the only subgroup of G is {e} , and {e} is generated by e. So we have the subgroups of G in the special case that G = {e} are cyclic. So we can assume G = where a ≠ e, and start the proof from there , and then it makes sense a^m is the smallest positive power of a in H.
You're the best maths teacher on UA-cam ❣️❣️
Thanks a lot Sushil! I've really been enjoying working on the abstract algebra lessons!
Thank you for explaining why r=0! My textbook didn't explain that part. Also, I like that we can see you in the video. I don't know why, but it helped me understand what you were saying.
awesome , well explained.
What if you started with G = , which implies that G = {e} since e^n = e for any integer n. Then it is trivial to show that any subgroup of G is cyclic, since the only subgroup of G is {e} , and {e} is generated by e. So we have the subgroups of G in the special case that G = {e} are cyclic.
So we can assume G = where a ≠ e, and start the proof from there , and then it makes sense a^m is the smallest positive power of a in H.
IDK how I got here or what I just watched, but I'm here for it.
Thank you so much! It's really helpful and your explanation is very easy to understand.
So in Euclid's Division Lemma, t can be negative, it's just that m has to be positive?
Yes or vice-versa, it's all about dividing by either positive or negative values, and r always being >=0 and less than m (which is positive)
Do you teach at uni in US?
Which apps you use?
Notability on iPad Pro!
Nothing round about with Wrath of Math! Yay!
It has been great to cycle back to some Abstract Algebra topics!