Every Subgroup of a Cyclic Group is Cyclic | Abstract Algebra
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- Опубліковано 22 сер 2024
- 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.
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Thanks a lot Sushil! I've really been enjoying working on the abstract algebra lessons!
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.
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.
So in Euclid's Division Lemma, t can be negative, it's just that m has to be positive?
Nothing round about with Wrath of Math! Yay!
It has been great to cycle back to some Abstract Algebra topics!
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Notability on iPad Pro!
Do you teach at uni in US?