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 p-subgroups of H $\displaystyle n_H$ is less thanor equal to the number of sylow p-subgroups 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 p-subgroup of H is a sylow p-subgroup 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 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.

m_s_d