Inner & Outer Semidirect Products Derivation - Group Theory

Semidirect products are a very important tool for studying groups because they allow us to break a group into smaller components using normal subgroups and complements! Here we describe a derivation for the idea of semidirect products and an explanation of how the map into the automorphism group relates to conjugation.
Group Theory playlist: • Group Theory
0:00 Inner semidirect product
11:18 Outer semidirect product
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Music: C418 - Pr Department

[@ProfOmarMath]

You've mathematically matured after a semester of college. It's really wonderful to see.

[@MuPrimeMath]

Thank you for the kind words!

[@user-xm9xo7jg4u]

This is the best lecture on semidirect products of groups I have ever seen. Thanks so much for your sharing.

[@maximpushkar91]

Very cool video! When this topic was on my lessons, I didn't really understand it. But now I can fully understand semidirect product, thank you!

[@guohaoyang9163]

This is the clearest video to introduce semi direct products! Thank you so much!🎉

[@ahyunseo]

What a perfect lecture

[@Astronautakaty]

this is truly perfect, thank you so much for this clear lecture

[@dipenganguly2635]

Nice video.. crystal clear

[@ayabarre9315]

Thanks...we've never taken this idea before this way❤

[@adresscenter]

Very good teacher

[@leeholzer4989]

that was SO helpful

[@sbjnyc]

I always wondered why the map is K-->Aut(H) rather than Inn(H) ... Can you make a semi direct product using outer automorphism?

[@sambhusharma1436]

Very nice 👍

[@artsygirlpoulami]

Thanks for such a nice and methodical explanation. Can you suggest to me a book regarding this abstract group theory part? I am facing difficulty in my course work.

[@MuPrimeMath]

The book that I used was Dummit and Foote. I didn't reference it very often so I don't actually know how helpful it is! However, it does have many group theory exercises for practice.

[@artsygirlpoulami]

@@MuPrimeMath Thanks a ton. I shall have a look.

[@awaiskhan8327]

How do we know that h1 = h2 and k1 =k2 are we assuming that each element in the group G, H is it's own inverse ?

[@tomkerruish2982]

Gotta say, you remind me of a young Allen Knutson, except that you're probably taller and I don't know if you juggle.

[@alexandrafant5735]

When you’re showing its a homomorphism, how is it not assuming the goal? I thought phi((h1,k1)(h2,k2))=phi((h1,k1))phi((h2,k2)) was the goal to show its a bijection, but in the end you use it as a fact to justify the products solution? I think I’m missing something

[@alexandrafant5735]

Ohhh nvm, we’re finding a law that makes it true! I get it now