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.
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.
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.
@@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.
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?
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.
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.
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
What a glorious response, thanks a lot for the elaboration! :)
Thanks a lot for taking the time out to write out such a detailed explanation.
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.
at 8:15, a small typo, G/ should be G/
At 18:18, I was wondering if you mean D_8 in the example.
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.
@@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.
Yes, this was a misprint for D_8
@@richarde.borcherds7998 Thank you for the clarification!
I don't think I followed some parts of the proofs. Could you tell me from which source they come from?
The book "algebra" by Lang contains proofs of most results in these lectures.
In Artin proofs look easy, but has some foundation before arriving at actual proof
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?
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.
Yo what's up Aakash lmao