Every Subgroup of a Cyclic Group is Cyclic | Abstract Algebra

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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[@Cowleotard]

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.

[@SushilKumar-su7oi]

You're the best maths teacher on KZbin ❣️❣️

[@WrathofMath]

Thanks a lot Sushil! I've really been enjoying working on the abstract algebra lessons!

[@Nayem_Uddin1]

Thank you so much! It's really helpful and your explanation is very easy to understand.

[@maxpercer7119]

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.

[@MarkWeaver-w2l]

IDK how I got here or what I just watched, but I'm here for it.

[@MrCoreyTexas]

So in Euclid's Division Lemma, t can be negative, it's just that m has to be positive?

[@punditgi]

Nothing round about with Wrath of Math! Yay!

[@WrathofMath]

It has been great to cycle back to some Abstract Algebra topics!

[@VS-is9yb]

Do you teach at uni in US?

[@__MonowarHossain]

Which apps you use?

[@WrathofMath]

Notability on iPad Pro!