Cosets -- Abstract Algebra video 9

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[Bedoroski]

This is some immense dedication. May it be a blessing that you can continue the great work

[harryh]

26:33 The subtle take-away from this video... "Parallelity" a noun that describes the state or condition of being parallel... Its actually a word, omfg!😂

[hawk]

Thank you so much for making this series!

[M.athematech]

The index of a subgroup is defined as the cardinality of the set of cosets of the subgroup. So yes it can be infinite, but you can't just use the sideways 8 infinity symbol, you have to distinguish between different infinite cardinals.

[schweinmachtbree]

The remark from 21:38 to 22:23 is wrong and would be better cut out from the video. xy^(-1) ϵ H is _not equivalent_ to x^(-1)y ϵ H. Note that H is closed under inverses, but when we invert x^(-1)y we get y^(-1)x by the shoes-and-socks lemma, not xy^(-1). The general situation is that for _left_ cosets, the "(5)-like" condition has the inverse on the _left_, i.e. xH = yH if and only if x^(-1)y ϵ H, which is equivalent to y^(-1)x ϵ H, while for _right_ cosets the corresponding (5)-like condition has the inverse on the _right_, i.e. Hx = Hy ⇔ xy^(-1) ϵ H ⇔ yx^(-1) ϵ H.
So tl;dr: the first condition that Michael writes in the circle is _not_ equivalent to the conditions in the theorem, which is about left cosets; rather it is equivalent to the corresponding right coset condition Hx = Hy. (and as Michael notes, left cosets are in general different from right cosets)

[esmurxes]

I have a question, is there a kind of bracket for groups that are not Lie algebras to measure how communicative two elements are? like [x, y] = xyx^-1y^-1?

[ra-hulu]

Which video group actions on a set will be thought ?

[matheusjahnke]

25:36... I think I know another proof.... it relies on the fact that equivalence classes partition a set... I knew this before hand but I don't think the series commented it before, so here goes.
Given an equivalence relation ~ in Z, a equivalence class [a] of an element a is a set { x ∈ Z ∣ x ~ a }
∣|
The equivalence classes are disjoint:
suppose there are two equivalence classes A ∪ B ≠ ∅.
Say... consider x ∈ [a] ∪ [b]
x ~ a because x ∈ [a]
x ~ b because x ∈ [b].... since the equivalence relation is symmetric, we can write b ~ x
... hence b ~ a... but if that is the case, then every element xb ∈ B is related to a, so B ⊆ A, similarly you can argue A ⊆ B... so A = B

The equivalence classes union to the set itself:
The proof is analogue to the one using in the video.. in 28:29, but I will include it for completeness
∀ x ∈ Z, [x] ⊆ Z
so ∪[x] ⊆ Z
Conversely, note that, since ~ is a equivalence relation, it is reflexive: ∀ x ∈ Z => x ~ x, so x ∈ [x] ⊆ ∪[x] ⊆ Z
hence ∪[x] = Z

Ergo, the equivalence classes induced by an equivalence relation Z partition it.

Let's define a relation ~ such that x ~ y if and only if xH=yH... actually, remember from the video... all the way from 10:30 till 25:30, xH=yH is equivalent to x⁻¹ y ∈ H... we will use that to prove ~ is equivalent

it is reflexive:
xx⁻¹=e ∈ H => x~x
it is symmetric:
x ~y => x⁻¹ y ∈ H => (x⁻¹ y)⁻¹= y⁻¹ x ∈ H => y ~ x
it is transitive:
x ~y and y~z => x⁻¹ y, y⁻¹ z ∈ H=> (x⁻¹ y)(y⁻¹ z) ∈ H=> x⁻¹z ∈ H => x ~ z

Guess what [x] = { y ∈ G| y ~ x } is. [x] = xH
y ~ x => x ~ y => xH = yH... but this implies(remember the list of equivalent statements) z ∈ xH
note that, every one of these steps is actually a biconditional, so you can go back from z ∈ xH and arrive at z ∈ [x]
so [x] ⊆ xH and xH ⊆[x]... namely xH = [x].... which we know partition G