I'm trying to show that if H is a subgroup of a finite group G, and p is a prime that the number of sylow p-subgroups of H is less thanor equal to the number of sylow p-subgroups of .
This is all I have:
Clearly if p does not divide |G| then p does not divide |H| and so . Let and be the greatest powers of p that divides |H| and |G| respectively.
Case 1: j = k. In this case any sylow p-subgroup of H is a sylow p-subgroup of G and so clearly .
Case 2: j < k.
This is where i got stuck. I am at a point where I need to show that each sylow p-subgroup of G can contain only one sylow p-subgroup of H. I do not know if this is right path, maybe I overlooked a simple theorem.
Any assistance is greatly appreciated. Thank you.