Kernels of 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

[@MrCoreyTexas]

Some things were left unsaid. Fancy F(Fancy R) is a group because:. Elements would be things like 2x, x^2, 2^x, e^x, and so forth. The identity would be f(x)=0 and if you add f(x) and g(x) you get another function that's in that group. The inverse is of f(x) is simply -1*f(x), since if you add those you get 0. So things like sin(x) would also be group elements because sin(x) + -sin(x) =0. Inverse here is different than the usual inverse you think of from high school algebra. What was throwing me off is in high school algebra, the inverse of sin() is only defined on a certain set of numbers, whereas here, in this group context, the inverse of sin(x) is just -sin(x). Math has this problem in that there are only so many words and alphabetical characters that things get reused.

[@okuhlemene709]

Can you please do a video of Group Action(The orbits and Stabilizers).🙏

[@WrathofMath]

Sure! Just need some time to prepare it.

[@like_that4966]

​@@WrathofMath Continue on ring theory too please

[@MrMegatherium]

I am just starting out. Why do we care about this object x a x inverse.... this conjugation. What does it mean? Why do we care if it is closed under conjugation any way?

[@WrathofMath]

Short answer to your question is to watch the videos around this one in my playlist: kzbin.info/aero/PLztBpqftvzxVvdVmBMSM4PVeOsE5w1NnN Closure with respect to the conjugate implies the subgroup is normal, and we can take quotient groups by normal subgroups. Quotient groups we later show are essentially equivalent to homomorphic images (that's the Fundamental Homomorphism theorem).

[@MrCoreyTexas]

Don't think you have showed that f(abc)=f(a)f(b)f(c) but it seems like it's not too hard to prove