- #1
catcherintherye
- 48
- 0
okay so I'm going through the proof of sylow part 2, i.e. the bit that says if
[tex] N_p is the number of subgroups of G of order p^n then N_p \equiv1modp
[/tex]
now I have got to the part where I have taken the subgroup P of order P^n that you get from sylow part 1 and I have to show that it is a unique fixed point W.r.t conjugation of P on S={Q: Q subgroup of G: |Q|=p^n}...anyway for those of you who know the proof, I am trying to find the order of the subgroup PQ, where Q is the supposed subgroup used in the contradiction argument...
so anyway |PQ| is what I am trying to deduce, I have noethers 1st isomorphism theorem
[tex] PQ/Q isomorphic to P/P \capQ Q [/tex]
|P|=p^n...
the next line asserts [tex] |P\cap Q| =p^e for som e e\leq n [/tex]
how is this so is P intersection Q a subgroup of P??
[tex] N_p is the number of subgroups of G of order p^n then N_p \equiv1modp
[/tex]
now I have got to the part where I have taken the subgroup P of order P^n that you get from sylow part 1 and I have to show that it is a unique fixed point W.r.t conjugation of P on S={Q: Q subgroup of G: |Q|=p^n}...anyway for those of you who know the proof, I am trying to find the order of the subgroup PQ, where Q is the supposed subgroup used in the contradiction argument...
so anyway |PQ| is what I am trying to deduce, I have noethers 1st isomorphism theorem
[tex] PQ/Q isomorphic to P/P \capQ Q [/tex]
|P|=p^n...
the next line asserts [tex] |P\cap Q| =p^e for som e e\leq n [/tex]
how is this so is P intersection Q a subgroup of P??