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Let $G$ be a finite group and $H < G$. Prove that $n_p(H) \le n_p(G)$.

Ok, so now i know that $n_p(H) \le n_p(G)$ refers to the number of Sylow p-subgroups in H and G, respectively. From here, I could state that for any $P\in$ Syl $p(G)$ there is a conjugate $gPg^{−1}$ such that $gPg^{−1}\cap H \in$ Syl $p(H)$ for any subgroup H of G whether normal or not.

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    $\begingroup$ I think it is the number of $p$-Sylows because if it were about normalizer it would need a sentence after explaining that you must verify this up to conjugation (because the normalizer of a $p$-Sylow without giving the $p$-Sylow has no sense except if you are looking for the conjugacy class of the normalizer). $\endgroup$ – Clément Guérin Mar 25 '15 at 17:15
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Let $P \in Syl_p(H)$. Since $P$ is a $p$-group in $G$ it must be contained in some Sylow $p$-subgroup $S$ of $G$. Hence $P \subseteq H \cap S$, and in particular $|P| \leq |H \cap S|$. On the other hand, $H \cap S$ is a $p$-subgroup of $H$ and again, must be contained in some Sylow $p$-subgroup of $H$, say $P^h$, for some $h \in H$. Since $H \cap S \subseteq P^h$, we have $|H \cap S| \leq |P^h|=|P|$. We conclude that $|H \cap S|=|P|$, hence $H \cap S=P$ (and $h \in N_H(P)$). So with every $P \in Syl_p(H)$ one can associate an $S \in Syl_p(G)$, with $P=H \cap S$. This yields an injection of $Syl_p(H)$ into $Syl_p(G)$ and hence $n_p(H) \leq n_p(G)$.

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For $H<G$ first see that corresponding each sylow $p$ subgroup of $G$ we'll get a sylow $p$ subgroup of $G$. Then your proof follows right. (try to see it for $|G|=p^k n$ and $|H|=p^r n_1$ where $k>r$)

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