Some general results on groups -- Abstract Algebra 6

[@SeeTv.]

14:55 the given statement was used one line below.

[@jacoberu-q2w]

good writing, good explanations. you do an excellent job!

[@jeffbarrett2730]

Hi Michael, I love both your channels. Here's a more direct proof that the centralizer is a subgroup without first showing that inverses are in the group. Let a, b in C_G(S). Then consider sa = as sab^-1 = as b^-1 sab^-1 = a(b^-1 b)sb^-1 sab^-1 = (ab^-1) s (b b^-1) [since b in C_G(S)] And so we have s(ab^-1) = (ab^-1) s, so ab^-1 in C_G(S).

[@jomama3465]

Hello Mr. Penn. I have been watching your first channel since I was in high school and now that I'm taking BS Math, I find this other channel of yours to be extremely helpful. Thank you. I hope that you'll have a playlist here devoted to Advanced Calculus/Mathematicsl Analysis/Real Analysis.

[@schweinmachtbree1013]

The statement of the result at 15:25 is slightly wrong: not every element of a group can have order 2 because the identity always has order 1. you can fix it by writing either "If ord(g)=2 for all non-identity elements gϵG then..." or alternatively "If g^2 = e for all gϵG then..." (or with fancier terminology it can also be stated as "If G has exponent 2 then...")

[@matheusjahnke8643]

Another way would be ord(g) | 2 for all g... because then ord(g)=1 for g=e... and ord(g)=2 for everything else.

[@abdurrahmanlabib916]

Please update the diff eqq playlist

[@bjornfeuerbacher5514]

24:55 If G is finite and H is a non-empty subset, then it's enough to show that for all a,b in H, we have that also is in H, isn't it?! Because that implies that for all a in H, all natural powers of a also are in H. And that in turn implies that both the neutral element and the inverse element of every element also are in H.

[@Aman_iitbh]

Ryt

[@SeeTv.]

I'm stuck with the first warm uo question: Since g^10 = e, ord(g) divides 10, the possible candidates for ord(g) are 1,2,5,10. We can immediately see that 1 and 2 are impossible (they would both contradict g^2 ≠ e). But I can't find an argument as to why ord(g) is 5 or cannot be 5. (My first idea was to somehow use that g^3 and g^8 are a multiplication of g^5 apart, but I couldn't follow anything from this.) The cyclic groups G=Z_5 and G=Z_10 with addition modulo 5 or 10 respectively (identity = 0) and g=1 both satisfy all conditions but in one case the order of 1 os 5 and in the other it's 10. Is there not enough information or am I missing something?

[@seanbastian4614]

I agree. By the looks of it, the problem would have been a little more simpler if it said that for all b in G, b^2 is not the identity. Then g^5 could not be the identity or else e=g^10=g^5*g^5=(g^5)^2 would not be the identity. So the order would have to be 10.

[@SeeTv.]

@@seanbastian4614 Thank you, that makes sense.

[@Happy_Abe]

And how would one find that g^3 and g^8 aren’t the identity?

[@Happy_Abe]

Does it just follow from the order being 10?

[@SeeTv.]

@@Happy_Abe If a natural number n satisfies g^n = e, then (by a theorem) n must be a multiple of the order of g. That's it.

[@matheusjahnke8643]

CG(empty set) = NG(empty set)=G... because all values vacuously satisfy the conditions

[@epsilonator]

Hi Michael, love your lectures. Do you plan to make a series on Galois theory in the future (possibly building up to the proof of the Abel-Ruffini Theorem)?

[@lewistsao3279]

Is it not more usually the case that only left identity and left inverse is defined and it is then proven that these are the same as the right identity and right

[@aashsyed1277]

In which video do you talk about orders of a group and order of an element?

[@mathmajor]

video 3.

[@aashsyed1277]

​@@mathmajor thanks! I already found it before.

[@astriiix]

34:57 I might be wrong but i think {e, sr} and {e, sr²} are not groups: sr • sr = r², which is not in the group sr² • sr² = r, which is not in the group can someone correct me if im wrong?

[@mathmajor]

these are group. Think about it geometrically: sr and sr^2 are both reflections which have order 2. You can also check with the commutation relations: (sr)(sr)=(ss)(r^{n-1} r)=s^2r^n=e, and similar for the other.

[@valdemie4235]

you probably thought the composition was commutative, which is not, hence your mistake at your evaluations

[@StratosFair]

At 17:00 it would have been much quicker to just apply the result that was proven before. I mean it's not a big deal, but still, I'm lazy :p

[@SilverlightLantern]

30:37 LOL'd irl

[@andreasxfjd4141]

oh man, I immediately recognized: third semester in Germany 🙂

[@jorex6816]

Wie heißt das Modul, Algebra?

[@andreasxfjd4141]

@@jorex6816 ja

[@jorex6816]

@@andreasxfjd4141 Danke^^

[@Bodyknock]

40:20 It looks like something happened where the proof the Normalizer is a group accidentally got cut off

[@jorex6816]

That was probably on purpose because the definition of the normalizer was wrong in the first place

[@aashsyed1277]

😊 Also, first

[@Jeity_]

kzbin.info/www/bejne/ip-6qpuDpLmraas I think it's a detail, but it's important to use the non-emptiness of H in the proof of this theorem. All we showed here is that whenever a, b∈H, then there are some specific elements that are also in H, but this says nothing about the existence of them (in other words, all of these statements could be vacuously true). Using the fact that H is non-empty, then there is some h∈H, and because of the proposition in the theorem, then hh⁻¹ = e∈H, which then ensures that H has an identity.