
Sylow psubgroups
Hello,
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 psubgroups of H $\displaystyle n_H$ is less thanor equal to the number of sylow psubgroups of $\displaystyle n_G$.
This is all I have:
Clearly if p does not divide G then p does not divide H and so $\displaystyle n_H = n_G = 0$. Let $\displaystyle p^j$ and $\displaystyle p^k$ be the greatest powers of p that divides H and G respectively.
Case 1: j = k. In this case any sylow psubgroup of H is a sylow psubgroup of G and so clearly $\displaystyle n_H \le n_G$.
Case 2: j < k.
This is where i got stuck. I am at a point where I need to show that each sylow psubgroup of G can contain only one sylow psubgroup of H. I do not know if this is right path, maybe I overlooked a simple theorem.
Any assistance is greatly appreciated. Thank you.
m_s_d