Normal Subgroups and Quotient Groups -- Abstract Algebra 11

⭐Support the channel⭐
Patreon: / michaelpennmath
Merch: teespring.com/...
My amazon shop: www.amazon.com...
🟢 Discord: / discord
⭐my other channels⭐
Main Channel: / michaelpennmath
non-math podcast: / @thepennpavpodcast7878
⭐My Links⭐
Personal Website: www.michael-pen...
Instagram: / melp2718
Randolph College Math: www.randolphcol...
Research Gate profile: www.researchga...
Google Scholar profile: scholar.google...

[@schweinmachtbree1013]

There is a problem with left vs right cosets in this video - e.g. the equality of left cosets aH = cH is equivalent to the condition a^(-1)c ϵ H, not ca^(-1) ϵ H. So the argument on the board at 10:00 becomes correct when all the left cosets are changed to right cosets, where ca^(-1) ϵ H etc. is the correct condition - that is, change gH to Hg throughout.

[@matematicacommarcospaulo]

I liked very much the way you introduce normal subgroups in a constructive way... We don't see this approach, for instance, in the classic book of Herstein

[@noahtaul]

32:27 You’re not actually working with the quotient here, you’re working with the subgroup. 37:24 Same thing, your symmetry group is now , nothing to do with the quotient. To talk about the quotient, you need to describe characteristics of the square that are unchanged under the subgroup action. For instance, with the subgroup , the clockwise order of the sides is unchanged, and for the subgroup , the pair of parallel edges on top and bottom is unchanged, so the quotient groups act on those things instead of the square itself.

[@bjornfeuerbacher5514]

32:27 Why? He talks about only a single symmetry being left: reflection, i. e. s. So he is talking about the group {e, s}, which is the quotient group, not the subgroup. 37:24 Again: Why? The quotient group is {ebar, rbar}, with rbar being a rotation with rbar² = ebar, i.e. a rotation by 180°. The square he is talking about has _exactly_ that symmetry group, so he _is_ talking about the quotient group here. And is exactly isomorphic to the quotient group {ebar, rbar}, so I don't get why you claim that has nothing to do with the quotient. "To talk about the quotient, you need to describe characteristics of the square that are unchanged under the subgroup action." Which is essentially what he did do, by adding some features to the squares, so one can clearly see what is unchanged by the actions and what isn't.

[@AbuMaxime]

Thank you for making me more intelligent when the rest of the world tries to make me more dumb.

[@jordimartinez4295]

Are we going to see some Galois theory?

[@schweinmachtbree1013]

Very likely not, because these videos are supporting the undergraduate class Michael is teaching this semester, and it would be very difficult to squeeze both group theory and Galois theory into one semester. (one _can_ approach groups from a Galois-theoretic angle -- indeed this is how groups were discovered -- but if Michael were taking this approach we would've already seen some Galois theory by now)

[@bjornfeuerbacher5514]

42:00 Wouldn't it be more sensible to point out that this quotient group is isomorphic to D_2?

[@karl131058]

Sorry, I'm not sure, ist THIS valid: In your "not-well-defined" example, the representations on the left have a 2-cycle as representative, so they should be elements of order 2 in the "new" group, but on the right they are represented by 3-cycles, so they should be of order 3 - contradiction!

[@schweinmachtbree1013]

Your argument works but you have to be careful: Let C = (2 3)H = (1 3 2)H. The fact that C is represented by a 2-cycle and a 3-cycle means that in the new 'group' it has order (which I will denote by "Ord", to distinguish from "ord" denoting orders in the original group) _dividing_ 2 and _dividing_ 3, but then Ord(C) | 2 and Ord(C) | 3 ⇒ Ord(C) | gcd(2, 3) = 1 ⇒ Ord(C) = 1, which is a contradiction because C is not the identity in the new 'group' since C ≠ H. This generalises to a subgroup H of an arbitrary group G as follows: if we can find elements x and y in G such that xH = yH := C ≠ H and gcd(ord(x), ord(y)) = 1 then H is not normal because, if the quotient group G/H were well-defined, we would have Ord(C) | ord(x) and Ord(C) | ord(y) ⇒ Ord(C) | gcd(ord(x), ord(y)) = 1, contradicting that C ≠ H. Phrased not in terms of cosets, this is equivalent to finding elements x, y ϵ G ∖ H with x^(-1)y ϵ H such that ord(x) and ord(y) are coprime.

[@M.athematech]

The exposition of the proof of well-definedness of the multiplication of cosets is not correct because you make use of the operation in the proof before it is actually shown to be well defined.

[@bjornfeuerbacher5514]

Huh? If one wants to check if an operation is well-defined, obviously one _has_ to make use of that operation.

[@M.athematech]

@@bjornfeuerbacher5514 nope one has to make use of the statements in the definition

[@bjornfeuerbacher5514]

​@@M.athematech Err, yes - if one wants to check if an operation is well-defined, then one has to make use of how the operation is defined. Could you please point out exactly at what point he is doing something wrong, please? (give a timestamp)

[@M.athematech]

@@bjornfeuerbacher5514 from 16:43 he starts by multiplying two cosets ... yet we are still trying to prove that this is actually well-defined. One cannot use the definition until after one shows it is well-defined. If he dropped the first expression that already uses coset multiplication and the final one that uses it again, then all the intermediate steps are sufficient for the proof by themselves. And only then can one start using coset multiplication.

[@bjornfeuerbacher5514]

@@M.athematech "yet we are still trying to prove that this is actually well-defined" Yes, and that's precisely what he is showing in the next few minutes: he shows that (ac)N is the same as (bd)N. That what it _means_ for the operation to be "well-defined". "One cannot use the definition until after one shows it is well-defined. " His point is that he shows that (ab)N = (cd)N, by using the definition what (ab)N and (cd)N are supposed to be. Precisely by showing that these two are the same he shows that the operation is well-defined. You are merely squabbling about notation here, not about the actual mathematical logic he uses.

[@psychSage]

ваши размышления напомнили мне забавное высказывание: "математика - искусство называть разными именами одни вещи"

[@artemetra3262]

одни ваши что?

[@psychSage]

@@artemetra3262 исправлено