okay so I'm going through the proof of sylow part 2, i.e. the bit that says if(adsbygoogle = window.adsbygoogle || []).push({});

[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??

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Sylow theorem part 2

**Physics Forums | Science Articles, Homework Help, Discussion**