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!
I'm surely missing something, but how is this not (12)(23)(31)? Really new to transposition series, so please don't bite my head off if I'm being dense here.
@@AndyGoth111 Well, consider where 1 goes in your transposition series -- (12)(23)(31) first takes 1 to 3, which gets taken to 2, which gets taken to 1 again. Compare (123), which only takes 1 to 2, and you can see that they're not equivalent.
@@AndyGoth111 in the composition of (12)(23) , you act from the right transposition first. 3 goes to 2 by the right transposition, then 2 is acted on by the left transposition to be taken to 1. So overall, 3 goes to 1 by passing through both transpositions (first the right, then the left).
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.
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?
Video 6 will introduce vector spaces over an arbitrary field. All of this "set-up" allows us to look at vector spaces over an arbitrary field at a lower level than is standard.
I might be wrong, but this could work: th-cam.com/video/PvUrbpsXZLU/w-d-xo.html P.S. A piece of advice: make video 1.5 times faster, I speak very slowly)
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.
Bruh for some reason always thought you were a high school student
Am I the only one that saw the word "good" flash up on the screen at 17:23 or am I going mad?
@@Alex_Deam You’re not mad
32:45 Nice
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!
At 15:05 (12)(23) is the correct transposition series.
I'm surely missing something, but how is this not (12)(23)(31)? Really new to transposition series, so please don't bite my head off if I'm being dense here.
@@AndyGoth111 Well, consider where 1 goes in your transposition series -- (12)(23)(31) first takes 1 to 3, which gets taken to 2, which gets taken to 1 again. Compare (123), which only takes 1 to 2, and you can see that they're not equivalent.
@@PersonaRandomNumbers The part I'm missing is what happens to 3. Doesn't (123) take 3 to 1? Am I just making that up?
@@AndyGoth111 in the composition of (12)(23) , you act from the right transposition first. 3 goes to 2 by the right transposition, then 2 is acted on by the left transposition to be taken to 1. So overall, 3 goes to 1 by passing through both transpositions (first the right, then the left).
This is a correct alternative but the one Penn wrote is good as well
Thanks!
Thank you very much for those amazing format ! It is very helpful and well explained !
You forget a condition in subgroup test set is not empty
if the set is empty then it doesn't have the identity and is not a subgroup
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.
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?
that's right. you could also do (x^n)^-1 as (x^-1)^n = (x^n)^-1.
We all need closure at some point in our lives.
15:04 Isn't that (1 2)(2 3).
Edit: Got rid of the commas.
Wait no I'm wrong dammit. its (1 2)(1 3)
I am just waiting to when we really get into the meat of things.
Video 6 will introduce vector spaces over an arbitrary field. All of this "set-up" allows us to look at vector spaces over an arbitrary field at a lower level than is standard.
I might be wrong, but this could work: th-cam.com/video/PvUrbpsXZLU/w-d-xo.html
P.S. A piece of advice: make video 1.5 times faster, I speak very slowly)