Visual Group Theory, Lecture 4.5: The isomorphism theorems

preview_player
Показать описание
Visual Group Theory, Lecture 4.5: The isomorphism theorems

There are four central results in group theory that are collectively known at the isomorphism theorems. We introduced the first of these a few lectures back, under the name of the "fundamental homomorphism theorem." In this lecture, we will see the other three, and we will motivate each one visually. We will prove one of these, sketch the proof of another, and leave the last proof as an exercise. After that, we introduce the notion of a commutator, which corresponds to a "non-abelian fragment" of a Cayley diagram. The commutators generate a normal subgroups, whose quotient yields an abelian group. We state this as a universal property, and close with a few examples.

Рекомендации по теме
Комментарии
Автор

As others have noticed, in the second isomorphism theorem the intersection of H and N need not be normal in G. As a counterexample, consider G = D_8, N = {e, r^2, s, sr^2}, H = {e, s}.

Jkfgjfgjfkjg
Автор

Error: in the second isomorphism theorem, [...] is a normal subgroup of H, not G.

We know that we may have H < N < G where N is normal in G and H is not, IE, normality isn't transitive in this way. Any instance of this serves as a counterexample to what's written.

EpicMathTime
Автор

36:40 One thing to be careful: Commutator elements of the form aba^-1b^-1does not form a subgroup of G. Rather, they generate the commutator subgroup of G.

sahhaf
Автор

Thank you Professor. I was stuck on one page for days. You made it so easy. I wish I could be in your classes.

saquibmohammad
Автор

In second isomorphism theorem the statement H intersection N is a normal sub group of G is wrong. It should be H intersection N is a normal subgroup of H.

hrithikchopra
Автор

Why is (H intersect N) normal in G? Clearly any element in it when conjugated by g in G lies in N (since N is normal), but why is it also in H? That part kind of gets ignored in the video. Thanks!

chetedoherty
Автор

@45:45 Those are not the only normal subgroups of D_4. The other subgroups of index 2 are normal as well.

Jkfgjfgjfkjg
Автор

You're a great teacher- enjoy your lectures!

oozecandy
Автор

Thanks for the uploads! They've been extremely helpful!

kansterstrak
Автор

There are many typos in the slides.. For example, in these slides (ii) of the second isomorphism theorem is wrong. The author would do a great service if he edits the videos and corrects the typos.

sahhaf
Автор

Excellent explanation Professor. Thank you.

yeisonquiceno
Автор

5:42 The idea is to paint a landscape where the proof is obvious...Pierre Deligne

smackronme
Автор

2nd theorem: Can't get to H-int-N being normal in G, but I can prove it in H. Seems like that works OK because you only need H/(H-int-N) rather than G/(H-int-N). Is that right?

rasraster
Автор

I actually said “wow” out loud after the demonstration of why abelianization works

michaelking
Автор

This lecture sounds noticeably more difficult/less clear than others.

kook
Автор

Please provide a proof that commutator subgroup is a normal subgroup

surensingh