GT6. Centralizers, Normalizers, and Direct Products

Abstract Algebra: We consider further methods of constructing new groups from old. We consider centralizer and normalizer subgroups, which are useful when the group is non-abelian, and direct products.

[MikkoHaavisto]

I was getting behind on algebra because of missing lectures, but this 84 video playlist saved me :) I wish universities invested if your efforts and instead of different lectures for every year, you would provide the video lectures, which can be reused.

I like the efficient teaching style you got. No time wasted, because you don't say anything redundant.

[popularmonster]

8 years since this video came out and still is useful nowadays, maths never get old!

[MathDoctorBob]

You're welcome! One of my first subscribers! Good luck with exams. Let me know if you have questions on anything. - Bob

[MathDoctorBob]

You're welcome! The most efficient way to describe a subgroup is as a nonempty subset of a group that is also a group. Efficiency isn't great for good teaching. It tends to bury the "why".

[CrackMeBros]

Watching your classes from France, that's really interesting, thanks for all Doctor 👍

[MathDoctorBob]

Set up multiplication tables. One element must be the identity. For the remainder, there are either three elements of order 2 (self-inverse), or an element of order 2 and an inverse pair. That should get you started.

[xiyongyan]

best presentation --Dr. Bob
Thank you

[jimnewton]

Hi Dr Bob. I studied group theory 25 years ago, but all of a sudden I need it again when working with Kleene algebras. You're videos are a great review. Thanks.

BTW do you know if there is a version of the fundamental homomorphism theorem for monoids? my guess is no because you need inverses to have cosets and without cosets you have no quotient group. But maybe there is something to salvage?

[ymgabaldon]

Thank you very much for your videos from Spain!! I'm about to start studying maths and they are helping me a lot. Only a suggestion. Why you don't use to prove if a subset is a subgroup that if x, y belong to H, H is a subgroup iff xy-1 belongs to H? That's the way I usually find in my book and I think is quite direct. Anyway, muchas gracias. :-)

[everything_strength]

How do you show that there are only 2 groups of order 4?

[izzgut]

what kind of jobs require this mathimatical theory

[ninosawbrzostowiecki]

I wish this guy was replaced by Liliana Castro, if we could only apply the replacement theorem from linear algebra to abstract algebra videos about groups...