Visual Group Theory, Lecture 4.5: The isomorphism theorems

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.
Course webpage (with lecture notes, HW, etc.): www.math.clemso...

[@EpicMathTime]

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.

[@jalilmamayo7642]

Nothing wrong in that part. N should really be a normal subgroup of G and not necessarily in H.

[@MuffinsAPlenty]

@@jalilmamayo7642 Because of EpicMathTime's use of an ellipsis, I think an ambiguity has arisen in EpicMathTime's comment. It seems you interpreted EpicMathTime's comment to be about the hypotheses of the theorem: that we take N ⊲ G. And it seems you interpreted EpicMathTime's comment as saying that the correct hypothesis is to take N ⊲ H, instead of G. You are correct that the hypotheses of the theorem are fine. And, indeed, if one takes N ⊲ H as a hypothesis, then the Second Isomorphism Theorem is a triviality, since in such a case, HN = H and H∩N = N. However, one could also interpret EpicMathTime's comment as saying conclusion (ii) is incorrect: the correct conclusion is that H∩N ⊲ H (rather than saying H∩N ⊲ G). And EpicMathTime's comment here is correct. As an example, consider D₄ the dihedral group of order 8. Let's call the generators r and s so that r⁴ = s² = 1 and rs = sr⁻¹ (i.e., r is a 90 degree rotation and s is a reflection). Let N = {1, s, r², sr²}. Then N ⊲ D₄ (it has index 2). Now consider H = {1, s}. One can check H < D₄, but that H is not _normal_ in D₄. Indeed, rsr⁻¹ = sr², which is not in H. But H∩N = H. So this is an example of a situation satisfying the hypothesis of the Second Isomorphism Theorem, but in which conclusion (ii), as written, is false.

[@Jkfgjfgjfkjg]

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}.

[@sahhaf1234]

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.

[@saquibmohammad2860]

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

[@hrithikchopra]

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.

[@12345shipreck]

No, its right they are equivalent. H intersection N is a normal subgroup of H, which is a subgroup of G. So H intersection N is a normal subgroup of G also.

[@Jkfgjfgjfkjg]

@@12345shipreck No it's not, because normality of subgroups if not transitive.

[@Jkfgjfgjfkjg]

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

[@chetedoherty]

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!

[@EpicMathTime]

It's not normal in G, and counterexamples are easy. It's normal in H.

[@AlexTheProMacGamer]

@@EpicMathTime Wow, I know it's an old comment, but thank you so much. I've been trying to prove something which is actually incorrect lol

[@oozecandy]

You're a great teacher- enjoy your lectures!

[@yeisonquiceno6747]

Excellent explanation Professor. Thank you.

[@sahhaf1234]

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.

[@kansterstrak]

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

[@smackronme]

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

[@michaelking7701]

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

[@rasraster]

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?

[@antoinelecalvez8]

R asraster Actually H-int-N has be a normal subgroup of H so H/H-int-N can be defined. But if H-int-N is a normal subgroup of H, since H

[@Jkfgjfgjfkjg]

@@antoinelecalvez8 This is unfortunately not true, as normality is not transitive. Meaning, if H is normal in K and K is normal in G, this does not imply that H is normal in G.

[@surensingh9114]

Please provide a proof that commutator subgroup is a normal subgroup

[@chuvzzz]

Note that his definition of the commutator subgroup is subtle. It is not made by the commutators themselves only; rather, the commutator subgroup is the defined as the set of all products of elements that have the form of a commutator (i.e. each individually is of the form abb*a*). With that, the proof that it is a subgroup is trivial, and so is the normality proof by inserting 1=g*g in between the a's and b's of gabb*a*g*

[@nm-de3bw]

​​@@chuvzzzabb*a*, not aba*b*?

[@kook6158]

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

[@pauselab5569]

it's a harder part, probably the most important theorems though especially because they prove Jordan Holder's theorem