Abstract Algebra | If G/Z(G) is cyclic then G is abelian.

We prove a classic result that says if the quotient of a group by its center is cyclic then the group is abelian.
www.michael-penn.net
www.randolphcollege.edu/mathem...

[@charlesleninja]

At 2:35 you needed to include the negative numbers as well when taking the union. When |xZ(G)| is finite you need not do this because (xZ(G))^-1 = (xZ(G))^n-1 where n = |xZ(G)| but when the order of the generator is infinite, you will never reach its inverse by repeating it. Otherwise great educationnal material, I'm going trhough Dummit & Foote's Abstract Algebra myself and did this exact problem some days ago. We have the exact same answer.
Edit: As a (funny?) result from this, we get G/Z(G) = {1} becasue Z(G) = G since G is abelian, so we will never get a non-trivial cyclic group if we quotient a group by its center, making my correction obselete in hindsight.

[@zappingabhishek4003]

at 3:40, we could add an extra explanation saying that x^n*y*x^m*z could be written as x^(n+m)*y*z because y belongs to Z(G), and hence y*a=a*y for every 'a' belonging to G. (I had a hard time getting this step).

[@hyperduality2838]

Subspace topology is dual to quotient topology.
Elliptic curves are dual to modular forms.
"Always two there are" -- Yoda.

[@revathymohans3053]

Really superb sir..thanks a lot for this video

[@Black_white_you]

thanks a lot!

[@esturion001]

Thanks

[@francineochoaf.160]

thaks a lot

[@annjulius4282]

Great video sir

[@juliangonzalez8997]

Thanks