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

[@sanitylost3114]

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

[@aleenajose9813]

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

[@Bedoroski]

Was a bit confused about how a subgroup is defined here compared to the updated list of videos. It took me a while to figure out these are all equivalent. Quite interesting to see how different perceptions of mathematics point to the same thing, the same as in topology

[@WrathofMath]

Very interesting indeed, and theses tests will definitely come in handy. Thanks for watching!

[@Jacob-ye7gu]

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

[@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

[@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... ✨

[@SaiDharahasReddyIndrakanti]

how come we are using the classic arithmetic operations on exponents, like are they valid for all type of group operation?. 15:16

[@Nasirmah]

I got lost when you started using power rules. How do we know (a^j)^-1 which means the inverse might be in G should be written as powers of a. Inverse should be different based on composition used.

[@Nasirmah]

Ok I finally understood. I think the confusion comes from that this power rule extends to any binary operation since you are doing j times. From the property which was already proven aka (a*b)^-1 aka the inverse is equal to a^-1*b^-1. So the above is equal to (a^-1)^j aka a inverse j times. This means a inverse by j times cancels out all of the a terms in a^i except the left overs are a compositions of i-j times.

[@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?

[@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.