Group theory 14: Sylow theorems

This lecture is part of an online mathematics course on group theory. It gives the proofs of the Sylow theorems about the Sylow p-subgroups: those of order the largest power of p dividing the order of a group.
Correction: Yenan Wang pointed out that at 18:18 D4 should be D8, the dihedral group of order 8.

[@nickhills428]

I am commenting just in case anyone else has found bits of this proof difficult. I was fine until 13:00...
At that point, we were considering the action of S on the set of p-Sylow subgroups by conjugation, the orbits of which are the singleton set {S} or sets size divisible by p. The conclusion was then that the number of conjugates of S (i.e. by a general element of G) was then 1 mod p. This is obvious but I found it difficult (as did the comment above) so:
The point is that the number of conjugates of S is just precisely a union of the orbits of the action of S. This union obviously includes the singleton set.
Then assuming S, T non-conjugate p-Sylow subgroups, we have 2 actions on the set of all p-Sylow subgroups: by S or by T.
Again, conjugates of S are just unions of the orbits which include the singleton under the first action but don't (by assumption about T) under the second. Hence 0=1 mod p.
For the final part, we have picked a maximal subgroup order p^{a} not contained in a p-Sylow subgroup. The normaliser of X, N(X), has index divisible by p since otherwise N(X) would have a p-Sylow subgroup S (of order p^{n}) and the same size argument as before would give X contained in S.
Hence we have the number of conjugates of X is divisible by p (since this is precisely the index of N(X) in G).
If Y is another subgroup (not equal to X) of order p^{a}, X cannot normalise Y since we obtain the subgroup XY of order p^{k}, k>a not contained in a p-Sylow subgroup. This is where we use the maximality of X which I couldn't immediately see.

[@ChairPencil]

Why is the number of conjugates of S equal to the sum of the size of the orbits of S acting by conjugation on the Sylow-subgroups

[@xxMADxxSCIENTISTxx]

What a glorious response, thanks a lot for the elaboration! :)

[@tadaiyoradima]

Thanks a lot for taking the time out to write out such a detailed explanation.

[@manassrivastava1048]

I am a big fan of yours Sir. I attended summer school at Berkley in 2018, took algebra class with Prof. Paulin. I came on the 9th floor to see you, saw your name outside your office, one of the most overwhelming feelings. I am very much interested in category theory and monster groups too! I am an undergrad in India planning to apply for graduate studies.

[@larryli2613]

at 8:15, a small typo, G/ should be G/

[@miglastoiglas]

I don't think I followed some parts of the proofs. Could you tell me from which source they come from?

[@richarde.borcherds7998]

The book "algebra" by Lang contains proofs of most results in these lectures.

[@JaspreetSingh-zp2nm]

In Artin proofs look easy, but has some foundation before arriving at actual proof

[@yenanwang7019]

At 18:18, I was wondering if you mean D_8 in the example.

[@meesdevries457]

There are two conventions for naming Dihedral groups. You are using the convention where D_n has n elements; the video uses the convention where D_n is the symmetry group of the n-gon.

[@yenanwang7019]

@@meesdevries457 I see. Thank you. The only thing is in the Dihedral Group Section, it seems to me that Professor Borcherds was using the number of elements conventison.

[@richarde.borcherds7998]

Yes, this was a misprint for D_8

[@yenanwang7019]

@@richarde.borcherds7998 Thank you for the clarification!

[@aakashparikh8538]

Around the 13 minute mark, your proof of the second part of sylow's theorem doesn't make sense to me. You consider the action of S on Q, the set of p-subgroups, and argue that the orbits of Q under this action are either S itself (size 1) or must have size divisible by p (you write p^k but only show, state, and use that these other orbits have size divisible by p). Sure, now you have shown that the number of such p groups, a.k.a the size of Q, is 1 mod p. But then you start saying that this means that S has 1 mod p conjugate p groups-- this doesn't follow unless I am missing something. Where have you shown conjugacy of S to any other group?

[@mathstrek]

The reasoning is similar. You let Q' be the set of all conjugates of S and let S act on this set Q'. Then again, there is a singleton orbit {S} and all other orbits have size divisible by p.

[@stephenhu2000]

Yo what's up Aakash lmao