Group Homomorphisms -- Abstract Linear Algebra 4

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[Karatemaci]

At 15:05 (12)(23) is the correct transposition series.

[lexinwonderland]

YES!! I'm so glad to see this channel get rolling! I can't wait for more abstract algebra concepts, especially if we can get into more generalized algebraic structures and/or how it connects with topology!

[goodplacetostop]

17:23 Good
35:22 I have to say I really enjoy this series. That reminds me the first months of my computer science degree.

[MrRyanroberson]

Question: what about a subgroup in which the identity is not an element yet there exists some element such that for all THOSE elements it acts as an identity? This can occur when a set has idempotent elements, and I'm not sure whether such sets formally qualify as groups, but try the tricomplex numbers and you'll see. Let i^2=j, i^3=1, associative and commutative, notice (i-j)/sqrt(3) = z satisfies: z^4 = -z^2; z^(4n+k) = z^k for n, k natural numbers.

[sciencewithali]

Thank you very much for those amazing format ! It is very helpful and well explained !

[synaestheziac]

Loving the group theory. For the last homework problem, I thought that we had only defined the exponent notation for positive exponents so far (x^n is x operated with itself n times). I assume the definition of negative exponents is just x^-n = (x^-1)^n, where n > 0–is that right?

[noumanegaou]

You forget a condition in subgroup test set is not empty

[lachenmann]

We all need closure at some point in our lives.

[abrahammekonnen]

15:04 Isn't that (1 2)(2 3).

Edit: Got rid of the commas.

[aaronharrell]

I am just waiting to when we really get into the meat of things.

[artificialresearching]

P.S. A piece of advice: make video 1.5 times faster, I speak very slowly)