The alternating group -- Abstract Algebra 12

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[@thomashoffmann8857]

6:30 I think the transpositions in the second example have reversed order

[@schweinmachtbree1013]

Some little mistakes: - the proposition at 19:29 only holds for n≥2, because we need the transposition (1 2) to define the map f. (for n = 0 and 1 we have A_n = S_n ⇒ |A_n| = n! instead of |A_n| = n!/2) - at 25:20 it should likewise be n≥2 (Michael is right that the alternating group is boring in the case n=2, but it is also quite boring for n=3 since |A_3| = 3!/2 = 3 ⇒ A_3 is cyclic) - michael misspoke at 34:16 - he said "A_4" but meant A_n

[@matheusjahnke8643]

At 47:05 the previous proof failed for n=4 because, at 44:51, Michael used 5 distinct elements: 1,2,3,4,5 The proof doesn't need to use precisely these numbers. But it must have a1, a2, a3, a4 and a5 distinct from each other. It can't happen in S4 because they are permutations over sets of 4 elements.

[@dogedev1337]

The fact that A_5 is simple is one of my favorite theorems in group theory. Are simple groups of Lie type something that's going to be covered on the upcoming Lie algebras course?

[@synaestheziac]

31:00 it’s easy enough to *check* these cycle computations, but is there a method for coming up with them in the first place? Like, how would I arrive at (abc)=(12a)²(12c)(12b)²(12a)?

[@matheusjahnke8643]

He did before: (2 a b)=(1 2 a)²(1 2 b) if we exchange a with b, we have: (2 b a)=(1 2 b)²(1 2 a) If we exchange b with c in the original expression, we have: (2 a c)=(1 2 a)²(1 2 c) (2 a c)(2 b a)=(2)(a b c)=(a b c) (a b c)=(2 a c)(2 b a)=(1 2 a)²(1 2 c)(1 2 b)²(1 2 a) I assume you could eventually come up with them if you do a lot of exercises.

[@matheusjahnke8643]

Actually, at 13:10, just like a_(i+2) b_{i+2} might have changed, the element accompanying "a"... called "b" might have changed along the way, note the second equivalence at 11:30, which states (b c)(a b) = (a c)(b c)... in this case, we have b' = c. That doesn't change the argument except for putting a prime on b.

[@dalitlegreenfuzzyman]

THIS VIDEO IS SO SO HUGE FOR THOSE INTERESTED IN GALOIS THEORY

[@astriiix]

literally me rn, i watched a lot of videos on youtube and they all skip this step. like why dude this is literally why you dont have a quintic, sextic etc. formula which is the whole point of the video 😫

[@grigoriefimovitchrasputin5442]

For the asigment 4, is this a correct proof : let's say that n is not prime. We then have n = ab, where a and b are both different from 1. Let's take . Clearly, is order is b, and b

[@matheusjahnke8643]

You proved: "n not prime" => "Zn is not normal" "n is prime" => "Zn is normal" This is enough to prove: "n is prime" "Zn is normal" (note "n not prime" => "Zn is not normal" is equivalent to "n prime"

[@alegal695]

@grigoriefimovitchrasputin5442 I think the first line of your proof is incorrect. If n is not prime then you have two possibilitities: either n=1 , or n is composite. You included the "n= composite case" but you didn't mention the case n=1.

[@alegal695]

At 25:20: it should be n>=2

[@matheusjahnke8643]

at 44:50 the inverse should have been of cycle... not the inverse of 5. (1 3 5)⁻¹ instead of (1 3 5⁻¹)

[@Happy_Abe]

@30:18 why is this the last one to check? Wouldn’t we need to check all permutations? So (12a), (a12), (a21), (21a), (2a1), (a1b), (ab1), (a2b), and finally ((ab2) should also need checking. If not, why?

[@iabervon]

The permutations of a 3-cycle are either its square or just a different way to write it. That is, for all x, y, and z, (x y z) = (y z x) = (z x y), and (x z y) = (z y x) = (y x z) = (x y z)^2. (For any cycle, we can just rotate it, because the first element is implicitly after the last element.)

[@Happy_Abe]

@@iabervon ah makes so much sense, thank you very much!

[@schweinmachtbree1013]

Because the representations of cycles can be rotated as iabervon says, the only case which is not obvious is (2 a 1), but taking inspiration from (1 a 2) = (1 2 a)(1 2 a) one finds that (2 a 1) = (2 1 a)(2 1 a).

[@alegal695]

@@schweinmachtbree1013 the case (2 a 1) is easy: (2 a 1) = (1 2 a) by rotating the elements, and the latter is already in (1 2 something) form

[@Happy_Abe]

@15:33 why does (a,b) not having an inverse as we move it to the left mean there’s no other appearance of a?

[@iabervon]

We're always working on the rightmost appearance of a, wherever that is; by definition, there's no appearance of a to its right. The important point is that, after each step, there's still no a to the right of the new position of the a we were working on, so when it reaches the left, there must not be an a anywhere else.

[@Happy_Abe]

@@iabervon there’s something I’m not following here, yes we are always moving the rightmost appearance of a, but who’s to say there aren’t two or more two cycles that have a in them so we’ll be left with two appearances of a all the way on the left after moving them. All this was saying is that it doesn’t meet its inverse (a b) when being moved, why can’t there be an (a x) where x doesn’t equal b on the way so we end up with (a b)(a x)…after all the moving is done.

[@iabervon]

​@@Happy_Abe Ah, the last case in the observation is what happens if two transpositions involving a meet each other: you still get one more transposition without a on the right.

[@Happy_Abe]

@@iabervon I’m sorry for not understanding but I don’t see how this discounts the possibility of having two transpositions on the left of (a b)(a x) All we know is that (a b) doesn’t meet its inverse, how do we know there isn’t another (a x) somewhere else?

[@iabervon]

@@Happy_Abe If you have (a b)(a x)(stuff without a), that's fine, but you're not done yet. (a b)(a x)(stuff) = (a x)(b x)(stuff), and now you only have one transposition containing a. The rewriting operation we're doing can change the number of times a appears in the sequence, so we're not limited to shifting all of the "a" transpositions to the left end, we can combine them into a single "a" transposition.