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?
If you have the statement Forall g in G, Forall h in G, gh = hg then yes, the group is just abelian there. But that isn't what we have here We are looking at only the g which satisfy Forall h in G, gh = hg, its not always true for any g that we input we don't necessarily know that g^-1 satisfies even if g does, thats why we have to use the multiplication
I teach this course from: abstract.ups.edu/ That being said, this is a pretty standard result that could be found in most Abstract Algebra books/courses.
Nice mountains and even better examples , thank you.
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.
That’s what I’m wondering, how do we know that h * h_hat^-1 is in H?
Oh, H is already a subgroup. I see.
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?
If you have the statement Forall g in G, Forall h in G, gh = hg
then yes, the group is just abelian there.
But that isn't what we have here
We are looking at only the g which satisfy Forall h in G, gh = hg, its not always true for any g that we input
we don't necessarily know that g^-1 satisfies even if g does, thats why we have to use the multiplication
7:30, why you can use associativity? because C(H) is a group?
H is already a subgroup.
yuh!
What is the reference book of this video sir?
I teach this course from: abstract.ups.edu/
That being said, this is a pretty standard result that could be found in most Abstract Algebra books/courses.
What is your academic qualification?
Pancake seller?