The alternating group -- Abstract Algebra 12

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[thomashoffmann]

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

[schweinmachtbree]

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

[matheusjahnke]

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.

[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)?

[matheusjahnke]

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.

[AbuMaxime]

At around 24:51 it is not immediately obvious to me why An would be a normal subgroup.

[dalitlegreenfuzzyman]

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

[grigoriefimovitchrasputin]

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 <a>. Clearly, is order is b, and b<n, so <a> is a propre subgroup. However, since Zn is abelian, it follows that <a> is a normal subgroup (because for every g in <a> and h in En, hgh^-1 = hh^-1g = g ). Hence, if n is not prime, Zn obviously has a normal subgroup. So, since we're searching for Zn with no subgroups, we have to see what happens if n is prime because when n is not prime, Zn don't work. But it's a well known fact that the only subgroups of Zp are {0} and Zp himself (where p si prime), hence Zp don't have any proper normal subgroup because it don't have any subgroups other than {0} and himself . So, putting all of this together, we see that if n is a composite number, Zn necessary have at least a normal propre subgroup, and if n is prime, En don't have any proper normal subgroup non-trivial. Is this a valid proof?

[matheusjahnke]

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]

@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?

[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?