Two Step, One Step, and Finite Subgroup Tests | Abstract Algebra

All three subgroup tests in one video! We prove the one step subgroup test, the 2 step subgroup test, and the finite subgroup test! Each test is a way to show a nonempty subset of a group is a subgroup. We also do a two step subgroup test example, a one step subgroup test example, and a finite subgroup test example, showing the set various subsets of different groups are subgroups using these subgroup tests. #abstractalgebra
All About Subgroups: • All About Subgroups | ...
Order of Group Elements: • Order of Elements in a...
Finite Order Elements Have n Distinct Powers: • Proof: Finite Order El...
Abstract Algebra Course: • Abstract Algebra
Abstract Algebra Exercises: • Abstract Algebra Exerc...
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[@aleenajose9813]

Sir can you please explain graceful labeling and harmonious labeling of graphs.

[@sanitylost3114]

it took me 2 hours to understand this but it was worth it i guess

[@dineshkumare1750]

Wonderful Video!!!

[@WrathofMath]

Thank you! This video isn't even out yet, but you must be watching the Abstract Algebra playlist because you can see the finished but yet-to-be released videos if you use that. Let me know if you have any questions as you continue your studies!

[@dineshkumare1750]

@@WrathofMath Thank you so much.. So thankful to have a teacher like you. These Abstract Algebra playlist videos helps me a lot in my discrete structures paper... ✨

[@Jacob-ye7gu]

For proving inversion, I would just say a^n=e (where n>1), thus a^n-1 = a^-1.

[@angusclark6170]

There's two things I didn't understand about this video, mind you I'm self taught on this level of maths so there could very well be gaps in my knowledge (for example this is the first time I've heard of the Order of an element of a set). With the finite subgroup test, why does the group being finite imply that the order of x is finite? surely you can get cases where after transforming x a certain number of times you can reach a state that if you multiply by x another time you get the same state again? Like x^k = a x^(k+1) = ax = a? surely that's possible? then you can have an infinite order and a finite set?
And also for the finite subgroup test, when proving inverses exist, why can't you simplify it down to saying that if ord(x) = n then x^n = e = x^(n-1)*x therefore x^(n-1) is the inverse of x as they compose to make the identity and x^(n-1) is in H?

[@MrCoreyTexas]

Video 10 in this playlist depends on a concept (order) in video 13, so watching this in linear order got me confused! Watch video 13 first...

[@nyxiiestelle]

yeah same lol

[@sams_study_hub]

❤❤❤

[@WrathofMath]

Thanks for watching!

[@Channel-zb1fi]

@@WrathofMath 10:56: How is the order of some element of H always finite?
Why is it always true that for x in H, x^n = e, where b is int. Could x not be a value such that x^n never reaches e for any integer n.
Nvm, I get it.