Intro to Group Homomorphisms | Abstract Algebra

[@WrathofMath]

Support this course by joining Wrath of Math to access exclusive and early abstract algebra videos, plus lecture notes at the premium tier! kzbin.info/door/yEKvaxi8mt9FMc62MHcliwjoin Abstract Algebra Course: kzbin.info/aero/PLztBpqftvzxWT5z53AxSqkSaWDhAeToDG Abstract Algebra Exercises: kzbin.info/aero/PLztBpqftvzxVmiiFW7KtPwBpnHNkTVeJc

[@liketsontobo8463]

Socratica needs to watch out, your videos are way more clear than any I've seen

[@WrathofMath]

That’s a high compliment - thank you! Socratica really is the only competition I can see for well-produced abstract algebra videos, and they’re no longer active on that front. Michael Penn also occasionally makes some excellent algebra content.

[@Misanthropeee]

Thank you so much for these!! Been using you to study for finals next week ...

[@WrathofMath]

Awesome, I'm glad to help! Good luck next week!

[@yizhan_xzyibo]

Your vedios are really helpful and simple thankyou ❤️

[@WrathofMath]

Glad you like them!

[@OmodAkegna]

Very understandbly explanation

[@WrathofMath]

Thank you!

[@walker9904]

The Greatest of all times!

[@VijitChandna]

Right on time for my finals!

[@WrathofMath]

Awesome! Good luck!

[@punditgi]

Nice explanation! 😊

[@WrathofMath]

Thanks Ezra!

[@turtius]

this subject is an absolute nightmare, thanks for making it slightly easier to understand!

[@turtius]

@cjjk9142 a bit of exaggeration 😅

[@WrathofMath]

I'm doing my best! Good luck and let me know if you ever have any questions!

[@MrCoreyTexas]

It gets so abstract, it's easy to lose focus of the big picture, and get lost in all the a's,b's,p's and q's!. It's easier to think in concrete terms than in abstract terms (at least for myself, and many people).

[@revanthkalavala1829]

2) f(a).f(inv(a)) = f(a.inv(a)) = f(e_g) = e_h Hence f(inv(a)) is inverse of f(a)

[@MrCoreyTexas]

Which element is the identity element in the parity group introduced in the first example? It has to be e for even. I did a quick bing search and you can subsitute +1 for even and -1 for odd and let the group operation be multiplication. It's just a coincidence that 'e' stands for even and happens to be the symbol used as the identity for a general abstract group.

[@YoavTamari]

Very good!

[@WrathofMath]

Thank you!

[@Syrian.Coffee]

Wonderful video

[@WrathofMath]

Thank you!

[@teachMathonYoutube]

Example @11.37, the H group should be R instead of R* I think. Else -1 will not be included in H. Correct me if i am wrong.

[@jakebarnes6161]

R* just means the reals without 0. So it does include -1

[@MrCoreyTexas]

What was left unsaid by the video creator is, H has to leave out 0 because 0 doesn't have a multiplicative inverse, like 1/1000 has a mult. inverse of 1000, but there's nothing you can multiply 0 to get to 1. Question for big math geeks: Since they defined i as the square root of -1, why can't I define flibbleglobber to be the number you mulitply 0 by to get 1?

[@revanthkalavala1829]

1) f(e_g.e_g) = f(e_g) f(e_g) f(e_g) = f(e_g)f(e_g) Even after applying Group H ooeration, image diesnt change f(e_g) must be an identity element in codomain group H Hence,F(e_g) = e_h

[@VijitChandna]

Is the isomorphism video out yet? my exam is in 3 days ;_;

[@WrathofMath]

It comes out tonight! Here's an early link just for you! kzbin.info/www/bejne/nHjUqZJje5ythJY

[@user-wr4yl7tx3w]

Thanks!

[@ChocolateMilkCultLeader]

Wish you created this two weeks ago

[@WrathofMath]

Better late than never! Maybe in 7 years I'll have reached my goal of covering a full undergrad curriculum.

[@MrCoreyTexas]

Can anybody find a homomorphism that proves H is a homomorphic image of G in the Example 2?

[@user-wr4yl7tx3w]

Do we know why such math even came into existence? Seems like their could be practical applications.

[@MrCoreyTexas]

Guys in the 19th century were bored with books, and they had no internet or TV

[@aperson6389]

Based