Let \(G\) be a finite group whose order is divisible by a prime number \(p\), and let \(P \trianglelefteq G\) be a normal \(p\)-subgroup of \(G\) with \(|P| = p^c\) for some integer \(c \ge 1\).
Show that \(P\) is contained in every Sylow \(p\)-subgroup of \(G\).
Let \(M\) be a maximal proper subgroup of \(G\). Show that either \(P \subseteq M\) or the index satisfies \([G : M] = p^b\) for some integer \(1 \le b \le c\).
Hint
Normality of \(P\) buys you a subgroup \(PS\), not merely a subset. Compute its order.
Solution
Goal: Prove that a normal \(p\)-subgroup is contained in the intersection of all Sylow \(p\)-subgroups in (a), and deduce that a maximal subgroup either contains \(P\) or has \(p\)-power index in (b).
1. Part (a): \(P\) is contained in every Sylow \(p\)-subgroup of \(G\).
1.1. By the Sylow Theorems, \(G\) contains at least one Sylow \(p\)-subgroup \(Q_0 \in \operatorname{Syl}_p(G)\), and every \(p\)-subgroup of \(G\) is contained in some Sylow \(p\)-subgroup.
1.2. Since \(P\) is a \(p\)-subgroup of \(G\), choose a Sylow \(p\)-subgroup \(Q_0 \in \operatorname{Syl}_p(G)\) such that \(P \subseteq Q_0\).
1.3. Let \(Q \in \operatorname{Syl}_p(G)\) be an arbitrary Sylow \(p\)-subgroup.
1.4. By the Second Sylow Theorem, all Sylow \(p\)-subgroups of \(G\) are conjugate: there exists an element \(g \in G\) such that \(Q = g Q_0 g^{-1}\).
1.5. Conjugating the inclusion \(P \subseteq Q_0\) by \(g\) gives:\[g P g^{-1} \subseteq g Q_0 g^{-1} = Q.\]
1.6. Since \(P\) is normal in \(G\), \(g P g^{-1} = P\).
1.7. Thus \(P \subseteq Q\).
1.8. Since \(Q \in \operatorname{Syl}_p(G)\) was arbitrary, \(P \subseteq \bigcap_{Q \in \operatorname{Syl}_p(G)} Q\).
2. Part (b): Either \(P \subseteq M\) or \([G : M] = p^b\) with \(1 \le b \le c\).
2.1. Let \(M < G\) be a maximal proper subgroup.
2.2. If \(P \subseteq M\), the first alternative holds.
2.3. Suppose \(P \not\subseteq M\).
2.4. Subgroup property of the product: Since \(P \trianglelefteq G\) is normal, the set product \(M P = \{m p : m \in M, \, p \in P\}\) is a subgroup of \(G\).
2.5. Strict inclusion: Since \(P \not\subseteq M\), there exists an element \(p_0 \in P \setminus M\). Since \(e \in M\), \(p_0 = e p_0 \in M P\), which implies \(M \subsetneq M P \le G\).
2.6. Maximality of \(M\): Since \(M\) is a maximal proper subgroup and \(M \subsetneq M P \le G\), we must have \(M P = G\).
2.7. Index calculation: By the Second Isomorphism Theorem for groups (or the product formula \(|M P| = \frac{|M| |P|}{|M \cap P|}\)):\[[G : M] = [M P : M] = [P : M \cap P] = \frac{|P|}{|M \cap P|}.\]
2.8. Prime power divisibility: The intersection \(M \cap P\) is a subgroup of \(P\). Since \(|P| = p^c\), Lagrange’s Theorem implies \(|M \cap P| = p^a\) for some integer \(0 \le a \le c\).
2.9. Since \(P \not\subseteq M\), \(M \cap P \ne P\), so \(a < c\).
2.10. Setting \(b = c - a\), we obtain\[[G : M] = \frac{p^c}{p^a} = p^{c - a} = p^b,\]where \(1 \le b \le c\).
3. Conclusion:
$P$ lies in every Sylow $p$-subgroup, and any maximal subgroup not containing $P$ has index $p^b$ with $1 \le b \le c$.