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Let $p$ be a prime. Suppose $N$ is a normal subgroup of a finite group $G$. If $n_{p,G}$ is the number of Sylow $p$-subgroups of $G$, and $n_{p,N}$ is the number of Sylow $p$-subgroups of $N$, then $n_{p,N}$ divides $n_{p,G}$.

I've been working on this problem for a while now. I have shown that if $P_G \in Syl_p(G)$, i.e., $P_G$ is a Sylow $p$-subgroup of $G$, then $P_G \cap N \in Syl_p(N)$, and that if $P_N \in Syl_p(N)$, then $P_N = P_G \cap N$ for some $P_G \in Syl_p(G)$. In conclusion, there is a surjective function $f:Syl_p(G) \to Syl_p(N)$ defined by $f(P) := P \cap N$. Therefore, $n_{p,N} \leq n_{p,G}$.

However, I cannot prove the divisibility relation.

Any help will be appreciated.

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Hint: Do you recall Madam Noether's First Isomorphism theorem? Unrelated: Pondering about if the map is one-one clears up some ideas! – user21436 Feb 26 '12 at 1:20
up vote 3 down vote accepted

If we conjugate a $p$-Sylow subgroup of $N$ by an element of $G$, we get another $p$-Sylow subgroup of $N$, since $N$ is normal. If $I_G$ is the group of elements of $G$ who fix the $p$-Sylow sbgroups of $G$, and $I_N$ is the group of elements that fix the $p$-Sylow subgroups of $N$, then since restriction of a Sylow subgroup of $G$ is a Sylow sugrboup of $N$, we get a surjective homomorphism $G/I_G \to G/I_N$. Since all Sylow subgroups are conjugate, the result follows from the orbit-stabilizer theorem.

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