Fundamental Theorem of Abelian Groups -- Abstract Algebra 16

[@charliem6648]

0:00 Lemma: Let G be an abelian group with |G| = n and a prime p s.t. p/n. Then G has an element of order p
13:03 Lemma: G is a finite abelian p-group iff. |G| = p^n
16:33 Lemma: Let G be a finite abelian group. Then G is isomorphic to the cross product of its subgroups whose orders are the maximal powers of distinct primes (such that those powers of primes multiply to the order of G)
27:30 Lemma: Let G be a finite abelian p-group and take g as an element of G with maximal order. Then G is isomorphic to cross K, for some subgroup K of G
51:45 The Fundamental Theorem of Finite Abelian Groups
54:12 Examples

[@zadsar3406]

The first lemma is actually a corollary of Cauchy's Theorem, which says basically the same thing, but for any group, not just abelian ones. It's a very pretty theorem and the standard proof is inductive, building on this one in a pretty nice way. Also, there are some really beautiful proofs involving group actions and permutations.

[@valentinodrachuk5692]

Good to know. Thanks👍

[@Aaron-tt3jz]

@19:00
In an abelian group, the order(ab) is a divisor of the lcm(order(a),order(b)) but equality does not hold in general.
As a concrete example, consider the additive group Z_8.
Order(3) = 8 since (3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 0)
Order(1) = 8 since (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 0)
LCM(Order(3),Order(1)) = 8
However
Order(3 + 1) = Order(4) = 2 since (4 + 4=0)
(Note that the divisibility condition is all that's really needed in the proof.)

[@Mebasically]

39:37 For me this took a bit of time to figure out so fore anyone interested here's the proof for- r:=ord(gH)|ord(g)=:s :
We know-(g^r)H=H meaning g^r is in H.
Now we assume (for a contradiction) that r doesn't divide s. This means t:=gcd(r,s)

[@AbuMaxime]

Thanks. Very useful

[@scottmiller2591]

I followed along for a while, but at some point it just turned into waves of math speak. Definitely the most dense lesson so far.

[@schweinmachtbree1013]

Small things: at 6:25 Michael misspeaks and says "normal" but means "proper", and the lemma at 13:29 doesn't require abelianness (indeed it wasn't used in the proof) - including the word "abelian" actually makes the result incorrectly stated, because of course |G| = p^n doesn't imply that G is abelian (for example the quaternion group is a non-abelian group of order 8). If you want to include abelianness just so that the lecture is focused on only abelian groups, then the lemma needs to be stated as "For a finite abelian group G, G is a p-group |G| = p^n".
Big thing: unfortunately the proof of the lemma at 20:44 is incorrect - the mistake is a very common one: the generalisation from binary internal direct products G = N_1 × N_2 to arbitrary internal direct products G = N_1 × N_2 × N_3 × ... is not the "obvious one"; we still need every element of G to be expressible as a product of elements from the normal subgroups N_i, but the condition N_1 ∩ N_2 = {e} does not become N_i ∩ N_j = {e} for i ≠ j; the correct condition is that each N_i intersects trivially with _the product_ of the remaining subgroups, i.e. N_i ∩ (N_1 ... N_(i-1) N_(i+1) ...) = {e}. For example for an internal direct product G = N_1 × N_2 × N_3 we need N_1 ∩ (N_2 N_3) = {e} = N_2 ∩ (N_1 N_3) = N_3 ∩ (N_1 N_2), and the "obvious" condition N_1 ∩ N_2 = {e} = N_2 ∩ N_3 = N_1 ∩ N_3 is not sufficient.
One can see why N_i ∩ (N_1 ... N_(i-1) N_(i+1) ...) = {e} is the correct condition by looking at the external direct product: e.g. with three factors G ≅ G_1 × G_2 × G_3 we have that the copy of G_1, G_1 × {e} × {e} = {(g_1, e, e) | g_1 ϵ G_1}, not only intersects trivially with the copies of G_2 and G_3, {(e, g_2, e) | g_2 ϵ G_2} and {(e, e, g_3) | g_3 ϵ G_3}, but also intersects trivially with their product {(e, g_2, e)(e, e, g_3) | g_2 ϵ G_2, g_3 ϵ G_3} = {(e, g_2, g_3) | g_2 ϵ G_2, g_3 ϵ G_3}. The same kind of thing is true for internal direct products (sums) of vector spaces, where you can visualise things geometrically: *R* ^3 is an internal direct product of *R* × {0} × {0}, {0} × *R* × {0}, and {0} × {0} × *R* -- the x axis, y axis, and z axis -- because the x axis intersects trivially with the sum of the y axis and z axis, which is the y-z plane - similarly the y axis intersects trivially with the x-z plane and the z axis intersects trivially with the x-y plane.

[@kono152]

I find it unfortunate that there are mistakes like that that I couldn't possibly spot, it makes it quite demotivating knowing i'm not actually learning the correct stuff :/

[@mahmoudalbahar1641]

Many thanks for your great videos.

[@kushagrasinghal8209]

someone please help me with the atgument made just before 48:86 about internal and external direct products I dont see how its implied that that intersection is the identity.

[@kushagrasinghal8209]

I dont agree with that argument tbh

[@Happy_Abe]

@27:25 we still need to show the Gi’s are a normal subgroup to claim the isomorphism

[@spearraiden586]

Yeah, but all the Gi's are subgroups of a finite *abelian* group. Thus there are all normal.

[@Happy_Abe]

@@spearraiden586 right!
I know they are, I’m just saying to use the theorem he did the condition is normal subgroup so I think should be mentioned

[@ko-prometheus]

♓♓♓♓♓