Abstract Algebra | The subgroup test

We present a nice result that can be used to test whether or not a subset is a subgroup.

[SupriyoChowdhury]

Nice mountains and even better examples, thank you.

[nikitakipriyanov]

from 10:14: you've lost ĥ¯¹. Since y=g¯¹ĥg, then you've shown that y¯¹=g¯¹ĥ¯¹g, so xy¯¹=g¯¹hgg¯¹ĥ¯¹g=g¯¹hĥ¯¹g.

[ThePharphis]

Isn't it trivial (without any left or right multiplication) that is gh = hg then g^-1 h = h g^-1 ?


Since g is an arbitrary element, and it's inverse is another arbitrary element within the group. Therefore either one can serve as "g" in the original condition, gh = hg?

[Hateusernamearentu]

7:30, why you can use associativity? because C(H) is a group?

[ghallyarrahman]

What is the reference book of this video sir?