Every Subgroup of a Cyclic Group is Cyclic | Abstract Algebra
ฝัง
- เผยแพร่เมื่อ 30 ต.ค. 2022
- We prove that all subgroups of cyclic groups are themselves cyclic. We will need Euclid's division algorithm/Euclid's division lemma for this proof. We take an arbitrary subgroup H from our Cyclic group G, then we take an arbitrary element a^t from H. Certainly, all powers of a^t are in H, since H is closed. Then, it only remains to prove that all elements of H are in fact powers of a^t. #AbstractAlgebra
Lesson on Cyclic Groups, Generators, and Cyclic Subgroups: • Cyclic Groups, Generat...
Abstract Algebra Course: • Abstract Algebra
Abstract Algebra Exercises:
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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.
Thank you so much! It's really helpful and your explanation is very easy to understand.
You're the best maths teacher on TH-cam ❣️❣️
Thanks a lot Sushil! I've really been enjoying working on the abstract algebra lessons!
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.
So in Euclid's Division Lemma, t can be negative, it's just that m has to be 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!