302.4B: Solvable Groups 

Not every normal subgroup contains the commutator subgroup, but every normal subgroup whose quotient is abelian does.

I am highshcool math teacher in Koea. You are awesome in math. Your technic to explain is too good. I appreciate you. Thank you. 나는 한국 고등학교 수학교사입니다. 당신은 정말로 수학 천재입니다. 그리고 알기 쉽게 잘 가르칩니다. 항상 존경합니다. 감사합니다.

Remember, [G,G] is not just the subgroup consisting of all commutators, but is the subgroup generated by the commutators. There exist groups that have a commutator subgroup that contains more than just commutators, the smallest of these groups having order 96. Just being persnickety about your definition of the subgroup! Such a group is discussed in Rotman's 'An Introduction to the Theory of Groups,' exercise 2.43 page 34 of the 4th edition.

Yes, if N < G is normal and M < H is normal, then N x M < G x H is normal (in a direct product G x H, elements from G only act on elements of G and likewise for H, so this follows from the definition). However, if M < H is not normal, then N x M < G x H is probably not normal for the same reason.

Right; in the notation Z5 semicross Z2, the first factor Z5 is a normal subgroup - but not of Z2... rather, a normal subgroup of the semidirect product (i.e. D5).

And that Z_p normal subgroup would be the abelian simple group I referenced before right? Okay yes I am just a little rusty on some things, but this is "clicking". Good to see things in a different way. Thank you.

Sure: {1} is a maximal normal subgroup of an abelian group iff that group is Z_p for some prime p. By the fundamental theorem of abelian groups, every finite abelian group has some Z_p as a normal subgroup.

I think that the commutator subgroup is contained in the normal subgroup iff the quotient group is abelian, not for any normal subgroup, right?

Sure, any normal subgroup N < G for which the quotient G/N is not abelian would work. For instance, the normal subgroup S_3 x {1} < S_3 x S_4 has nonabelian quotient isomorphic to S_4.

For the dihedral group of the regular 5 gon there is the normal subgroup Zmod5 but what would be the semidirect product? Based on the video it seems like D5=Z5 semicross Z2 but Z2 isn't a normal subgroup of Z5 since the order of it doesn't divide the order of Z5. So what is the case then?

Thanks Matthew for your video! It explains stuff well. But I notice something might be wrong in your video. Since the group Z mod 12 and the group Z mod 6 have different operations, Z mod 6 is not supposed to be a subgroup of it ( unless you create an isomorphism ). check the video at 6:17

Okay I see. And as for direct products that are normal, is it true that the direct product of two normal subgroups is normal? What about the direct product of a normal and not normal subgroup? Makes sense to me that the direct product of abelian groups is abelian so that is an answer for the first question. But what about the other case? I feel rusty on thinking about everything we think about for groups but for examples that are direct products of groups. I'd like to work with them a bit more

Also, if the last normal subgroup save for the trivial group has to be abelian and maximal in the case of the JH theorem, doesn't that mean it must be simple? So for a group to be solvable does it have to have an abelian simple normal subgroup contained within it (or is itself an abelian simple group which also fits this requirement being a normal subgroup in itself)?

....was 4 though we had collections of cardinality

you're a legend tnk u

(forgive me for the many question but...) direct products don't rely on any group they are isomorphic to but the way we define a semidirect product is inherently related to a group they are isomorphic to? ZnXZm very easily defined as just the set of 2tuples (a,b) where a is from Zn and b is from Zm but a semi direct product is the set of 2 tuples (a,b) where...?

Can we ever have a normal subgroup of a group that isn't apart of any composition series that proves the group's solvability?

If ℤ₁₂ is solvable, and we can find one such descending path, does it mean that _every_ such path would work for ℤ₁₂? Also, can it also be said that ℤ₁₂ is solvable because we can write it as a direct product of cyclic groups? (e.g. ℤ₁₂ = ℤ₂×ℤ₂×ℤ₃ = ℤ₂×ℤ₆ = ℤ₃×ℤ₄; I simply took the Abelian groups that you wrote under the triangles in purple, and crossed them together)

Hello Professor Salomone, would you mind recommending some text books that proved the observation in 12:09? Thanks so much.

Q. Why are solvable groups so heavy? A. Because they have Abelian pieces inside them.

Trivial group is normal to every group, but clearly the commutator subgroup is trivial iff G is abelian. Thus here is another issue

So all abelian groups are solvable, but there are some solvable groups that aren't abelian like Dsub4. But we do know that of all groups that are not abelian those that are simple are unsolvable. Only the simple groups that are abelian are solvable. And for the maximality condition for the JH theorem that means that of all collections of normal subgroups that prove solvability of a group there is a unique maximal cardinality as we had with Zmod12 where the maximal cardinality for a collection

02:40 How does this "definition" relate to equations being solvable? (Perhaps this is explained in a different video in this series) 02:59 😁 09:28 😀

Hello. It's me. The poor in Abstract Algebra kid. I'd like to say two things : 1.) Does a perfect group exist ? 2.) Could you kindly do a video / direct me if there is one showing a proof that every solvable group has normal subgroups that are solvable. You mentioned this as a summary in the Quintic impossible at the end. Thanks.

@1:15 the word is "portentous".

you are my god

You misspelled portentous

Very based video

Nice video, horrible notation