Abstract Alg, Lec 2A: Group Examples, Modular Arithmetic, Sets & Functions, Composition of Functions

Abstract Algebra Course Lecture 2A: Group Examples, Modular Arithmetic, Sets and Functions (Function vs Relation), Composition of Functions (Function Composition)

(0:00) Lecture structure.
(2:04) Review group examples from Lecture 1 (integers, rationals, reals, complexes).
(3:38) Group Example: special functions under composition.
(13:41) Group Example: GL(2,R) (General Linear Group of nonsingular 2 by 2 matrices under matrix multiplication...nonsingular means invertible, which is equivalent to the determinant being nonzero).
(16:18) Group Example: Zn, the group of integers 0, 1, 2, ..., n-1 under addition modulo n.
(19:08) Modular arithmetic in Mathematica.
(22:05) Quick review of domain, codomain, image, one-to-one (injective), onto (surjective), invertible functions, composition of functions.
(27:10) Images (under functions) of unions and intersection and the idea of proof of the fact that the image of the intersection of two sets is a subset of the intersection of the images of each set.
(32:23) Composition of one-to-one and onto functions (and bijections) along with "socks-shoes" property of inverse functions.

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[shannonstich]

You are a fantastic professor. Thank you for posting this series.

[FernandoVinny]

A good way of thinking about calculations mod n is "take out all the multiples of n"

[finaltheorygames]

The problem is that people start teaching this at university when really you need to be taught the material starting in high school.

[plaustrarius]

This is pure gold I am so excited to watch the rest of this lecture series!!! I remember proving Fermat's little theorem in discrete math, modular arithmetic is so amazing

[maxpercer]

For the proof in 9:00 it looks like you are repeatedly applying the definition of composition "(f o g)(a) = f(g(a))", which can be interpreted as a mechanical rule for moving around parentheses -like a game of symbols.

Using the more compact notation (as used in the video) we have the same definition
(αß)(a) = α(ß(a)).

Then we want to show (αß)Γ = α(ßΓ).
Proof: for all a in the domain of Γ we have by rearranging parentheses
((αß)Γ)(a) = (αß)(Γ(a))=α(ß(Γ(a))) = α((ßΓ)(a)) = (α(ßΓ))(a).

[anshukumartiwari]

Sir, where can I get the lecture notes???

[zandorpx]

I had been attending a Algebra Course here in Brazil until the COVID-19 outbreak stoped the classes. Thanks for uploading you lectures, that way i can keep studying. God Bless you professor.

[SergioSotoQuintero]

What a nice lecture. Congratulations and thank you for showing such passion at 7:00 am in the morning.

[fadiadaghestani]

19:36
144 % 7 more than 10
7×10= 70 + x =144
X=74 = 14 [why not 4?]
14× 7 = 98
Same as 44%7
7 goes evenly to 42
Rem =2

[zacharysmith]

Do you happen to post the problem sets too?

[rajatbisht]

Thanks sir you have helped me pass my college exams.Sir your approach to mathematics have inspired me to study maths with enthusiasm.Real analysis video developed that approach of deriving theorem which once seemed useless, sir you are an inspiration.Thank you sir once again.