Intro to Group Homomorphisms | Abstract Algebra

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Wrath of Math

Wrath of Math

Күн бұрын

Пікірлер: 41
@WrathofMath
@WrathofMath 3 ай бұрын
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
@liketsontobo8463 Жыл бұрын
Socratica needs to watch out, your videos are way more clear than any I've seen
@WrathofMath
@WrathofMath Жыл бұрын
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 Socratica really is the only competition I can see for well-produced abstract algebra videos, and they’re no longer active on that front.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.
@yokiram-3706
@yokiram-3706 Ай бұрын
For all of you who are watching this video, if you want to understand a Homomorphism you are in the right place. in fact if you are really interested in group theory check out this guy's playlist and you will love it.... believe me when i say this i have been in many places
@Misanthropeee
@Misanthropeee Жыл бұрын
Thank you so much for these!! Been using you to study for finals next week ...
@WrathofMath
@WrathofMath Жыл бұрын
Awesome, I'm glad to help! Good luck next week!
@walker9904
@walker9904 Жыл бұрын
The Greatest of all times!
@yizhan_xzyibo
@yizhan_xzyibo Жыл бұрын
Your vedios are really helpful and simple thankyou ❤️
@WrathofMath
@WrathofMath Жыл бұрын
Glad you like them!
@punditgi
@punditgi Жыл бұрын
Nice explanation! 😊
@WrathofMath
@WrathofMath Жыл бұрын
Thanks Ezra!
@OmodAkegna
@OmodAkegna Жыл бұрын
Very understandbly explanation
@WrathofMath
@WrathofMath Жыл бұрын
Thank you!
@VijitChandna
@VijitChandna Жыл бұрын
Right on time for my finals!
@WrathofMath
@WrathofMath Жыл бұрын
Awesome! Good luck!
@turtius
@turtius Жыл бұрын
this subject is an absolute nightmare, thanks for making it slightly easier to understand!
@turtius
@turtius Жыл бұрын
@cjjk9142 a bit of exaggeration 😅
@WrathofMath
@WrathofMath Жыл бұрын
I'm doing my best! Good luck and let me know if you ever have any questions!
@MrCoreyTexas
@MrCoreyTexas 6 ай бұрын
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).
@YoavTamari
@YoavTamari 7 ай бұрын
Very good!
@WrathofMath
@WrathofMath 7 ай бұрын
Thank you!
@MrCoreyTexas
@MrCoreyTexas 6 ай бұрын
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.
@Syrian.Coffee
@Syrian.Coffee Жыл бұрын
Wonderful video
@WrathofMath
@WrathofMath Жыл бұрын
Thank you!
@IslamBelkacemi-y4w
@IslamBelkacemi-y4w 2 ай бұрын
in the non-example, you assumed that the operation is + ?
@user-wr4yl7tx3w
@user-wr4yl7tx3w Жыл бұрын
Thanks!
@VijitChandna
@VijitChandna Жыл бұрын
Is the isomorphism video out yet? my exam is in 3 days ;_;
@WrathofMath
@WrathofMath Жыл бұрын
It comes out tonight! Here's an early link just for you! kzbin.info/www/bejne/nHjUqZJje5ythJY
@teachMathonYoutube
@teachMathonYoutube 10 ай бұрын
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
@jakebarnes6161 9 ай бұрын
R* just means the reals without 0. So it does include -1
@MrCoreyTexas
@MrCoreyTexas 5 ай бұрын
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
@revanthkalavala1829 Жыл бұрын
2) f(a).f(inv(a)) = f(a.inv(a)) = f(e_g) = e_h Hence f(inv(a)) is inverse of f(a)
@shinkansen1907
@shinkansen1907 Ай бұрын
legend
@WrathofMath
@WrathofMath Ай бұрын
Appreciate you!
@MrCoreyTexas
@MrCoreyTexas 5 ай бұрын
Can anybody find a homomorphism that proves H is a homomorphic image of G in the Example 2?
@revanthkalavala1829
@revanthkalavala1829 Жыл бұрын
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
@aperson6389
@aperson6389 Жыл бұрын
Based
@ChocolateMilkCultLeader
@ChocolateMilkCultLeader Жыл бұрын
Wish you created this two weeks ago
@WrathofMath
@WrathofMath Жыл бұрын
Better late than never! Maybe in 7 years I'll have reached my goal of covering a full undergrad curriculum.
@user-wr4yl7tx3w
@user-wr4yl7tx3w Жыл бұрын
Do we know why such math even came into existence? Seems like their could be practical applications.
@MrCoreyTexas
@MrCoreyTexas 5 ай бұрын
Guys in the 19th century were bored with books, and they had no internet or TV
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