If $Q/N$ is a Sylow $p$-subgroup of $G/N$ then $Q$ contains a Sylow $p$-subgroup of $G$, and every Sylow $p$-subgroup of $G/N$ is such an image

Proof.

1. Since \(Q/N\) is a Sylow \(p-\)subgroup of \(G/N\), we can write \(|G/N| = p^k l\) where \(\gcd(p, l) = 1\), and \(|Q/N| = p^k\).

   We can then write \(|G| = p^n m\) with \(\gcd(p, m) = 1\), where \(n\geq k\) and \(l\mid m\).

   By the third isomorphism theorem, we have \[\frac{G/N}{Q/N} \cong G/Q\] and so \[\left| \frac{G/N}{Q/N} \right| = \frac{|G/N|}{|Q/N|} = \frac{p^k l}{p^k} = l\]

   and so \(|G/Q| = l\) where \((p, l) = 1\), and thus \[|G/Q| = |G| / |Q| = l \implies |G| = |Q|~l.\]

   We then have \[p^n m = |Q|~l,\]

   and since \((p, l) = 1\), it must be the case that \(p^n\) divides \(|Q|\).

   So take \(P \in \mathrm{Syl}(p, Q)\), a Sylow \(p-\)subgroup of \(Q\). Then \(|P| = p^n\), since \(p^n\) is the full power of \(p\) dividing \(|Q|\), and \(p^n\) is also the full power of \(p\) dividing \(|G|\). So \(P\) is a Sylow \(p-\)subgroup of \(G\) contained in \(Q\), which is what was asked.

   Note that \(Q\) itself need not be a Sylow \(p-\)subgroup: \(p^n\) divides \(|Q|\), but \(|Q| = |G|/l\) carries the factor \(m/l\) as well, so \(Q\) is a \(p-\)group only in the case \(m = l\).

2. Let \(P_N \in \mathrm{Syl}(p, G/N)\). By the subgroup correspondence theorem, \(P_N = H/N\) for some \(H\leq G\) such that \(N \subseteq H\).

   So choose \(P_H \in \mathrm{Syl}(p, H)\); the claim is that \(P_H \in \mathrm{Syl}(p, G)\) and that \(\frac{P_HN}{N} \cong P_N\), which exhibits \(P_N\) as the image of a Sylow \(p-\)subgroup of \(G\).

   We first have \(P_H \in \mathrm{Syl}(p, G)\). Factor the index through \(H\): \[[G : P_H] = [G : H]\,[H : P_H].\] Here \(p\) does not divide \([H : P_H]\), since \(P_H\) is Sylow in \(H\). And by the correspondence theorem \([G : H] = [G/N : H/N] = [G/N : P_N]\), which \(p\) does not divide either, since \(P_N\) is Sylow in \(G/N\). So \(p\) does not divide \([G: P_H]\), which makes \(P_H\) a maximal \(p-\)subgroup in \(G\) and thus a Sylow \(p-\)subgroup.

   We then have \(P_HN/N \leq P_N\), because \(P_H \leq H\) and \(N \leq H\) give \(P_H N \leq H\), so \(P_HN/N \leq H/N = P_N \leq G/N\).

   However, it is also the case that \(P_HN/N \in \mathrm{Syl}(p, G/N)\). This follows because

   1. \(P_HN/N \cong P_H/(P_H \cap N)\) by the 2nd isomorphism theorem, a quotient of a \(p-\)group, so it is a \(p-\)group.

   2. \(P_H \subseteq P_HN \subseteq G \implies p\) doesn’t divide \([G: P_HN]\), since \(P_H\) is also a Sylow \(p-\)group of \(G\) and thus has maximal prime power dividing \(\left| G \right|\).

   3. \(N \subseteq P_H N \subseteq G \implies [G/N : P_H N/ N] = [G: P_H N]\)

   Taken together, this says that \(P_H N/ N\) is a \(p\)-group and \(p\) doesn’t divide \([G/N : P_HN / N]\), so it is a maximal \(p-\)subgroup and \(P_HN/N \in \mathrm{Syl}(p, G/N)\).

   But since \(P_HN/N \leq P_N\) and \(\left|P_HN/N\right| = \left|P_N\right|\), we must have \(P_HN/N = P_N\) as desired.

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