Let \(G\) be a group of order \(105 = 3 \cdot 5 \cdot 7\), and let \(P \in \operatorname{Syl}_3(G)\), \(Q \in \operatorname{Syl}_5(G)\), and \(R \in \operatorname{Syl}_7(G)\) be Sylow 3-, 5-, and 7-subgroups of \(G\), respectively.
Prove that at least one of \(Q\) and \(R\) is normal in \(G\).
Prove that \(G\) has a cyclic subgroup of order 35.
Prove that both \(Q\) and \(R\) are normal in \(G\).
Prove that if \(P\) is normal in \(G\), then \(G\) is cyclic.
Solution
Goal: Prove normality of Sylow subgroups, existence of a cyclic subgroup of order 35, and deduce that \(G\) is cyclic when all Sylow subgroups are normal.
1. Part (a): At least one of \(Q\) or \(R\) is normal in \(G\).
1.1. By Sylow’s Theorems, the number \(n_5\) of Sylow 5-subgroups satisfies \(n_5 \equiv 1 \pmod 5\) and \(n_5 \mid (3 \cdot 7) = 21\). Thus \(n_5 \in \{1, 21\}\).
1.2. The number \(n_7\) of Sylow 7-subgroups satisfies \(n_7 \equiv 1 \pmod 7\) and \(n_7 \mid (3 \cdot 5) = 15\). Thus \(n_7 \in \{1, 15\}\).
1.3. Suppose for contradiction that neither \(Q\) nor \(R\) is normal in \(G\), so \(n_5 = 21\) and \(n_7 = 15\).
1.4. Count elements:- Each Sylow 5-subgroup has prime order 5, so any two distinct Sylow 5-subgroups intersect only in the identity \(\{e\}\). Together they contain \(21 \cdot (5 - 1) = 84\) elements of order 5.- Each Sylow 7-subgroup has prime order 7, so any two distinct Sylow 7-subgroups intersect only in \(\{e\}\). Together they contain \(15 \cdot (7 - 1) = 90\) elements of order 7.- Since 5 and 7 are distinct primes, no element can have both order 5 and order 7.
1.5. The identity, the elements of order 5, and the elements of order 7 give at least\[1 + 84 + 90 = 175 \text{ distinct elements in } G.\]
1.6. This contradicts \(|G| = 105\).
1.7. Thus \(n_5 = 1\) or \(n_7 = 1\), so at least one of \(Q\) or \(R\) is normal in \(G\).
2. Part (b): \(G\) has a cyclic subgroup of order 35.
2.1. By Part (a), \(Q \trianglelefteq G\) or \(R \trianglelefteq G\).
2.2. If one of two subgroups is normal, their set product is a subgroup. Thus \(H = Q R \le G\).
2.3. By Lagrange’s Theorem, \(|Q \cap R|\) divides \(\gcd(|Q|, |R|) = \gcd(5, 7) = 1\), so \(Q \cap R = \{e\}\).
2.4. The order of \(H\) is\[|H| = \frac{|Q| |R|}{|Q \cap R|} = \frac{5 \cdot 7}{1} = 35.\]
2.5. \(H\) is cyclic:- In \(H\), the number of Sylow 7-subgroups \(n_7(H)\) satisfies \(n_7(H) \equiv 1 \pmod 7\) and \(n_7(H) \mid 5 \implies n_7(H) = 1\), so the Sylow 7-subgroup \(R \trianglelefteq H\).- The number of Sylow 5-subgroups \(n_5(H)\) satisfies \(n_5(H) \equiv 1 \pmod 5\) and \(n_5(H) \mid 7 \implies n_5(H) = 1\), so the Sylow 5-subgroup \(Q \trianglelefteq H\).- Since \(Q, R \trianglelefteq H\) and \(Q \cap R = \{e\}\), \(H \cong Q \times R \cong \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z} \cong \mathbb{Z}/35\mathbb{Z}\).
2.6. Thus \(H\) is a cyclic subgroup of order 35 in \(G\).
3. Part (c): Both \(Q\) and \(R\) are normal in \(G\).
3.1. Let \(H \le G\) be the cyclic subgroup of order 35 from Part (b).
3.2. The index of \(H\) in \(G\) is \([G : H] = \frac{105}{35} = 3\).
3.3. Since 3 is the smallest prime dividing \(|G| = 105\), any subgroup of index 3 is normal in \(G\). Thus \(H \trianglelefteq G\).
3.4. In the cyclic group \(H \cong \mathbb{Z}/35\mathbb{Z}\), \(Q\) is the unique subgroup of order 5, and \(R\) is the unique subgroup of order 7.
3.5. Because \(Q\) and \(R\) are uniquely determined by their orders in \(H\), they are characteristic subgroups of \(H\): for every automorphism \(\sigma \in \operatorname{Aut}(H)\), \(\sigma(Q) = Q\) and \(\sigma(R) = R\).
3.6. Since \(H \trianglelefteq G\), conjugation by any \(g \in G\) restricts to an automorphism of \(H\).
3.7. Thus \(g Q g^{-1} = Q\) and \(g R g^{-1} = R\) for all \(g \in G\), proving that both \(Q \trianglelefteq G\) and \(R \trianglelefteq G\).
4. Part (d): If \(P \trianglelefteq G\), then \(G\) is cyclic.
4.1. Assume \(P \trianglelefteq G\). By Part (c), \(Q \trianglelefteq G\) and \(R \trianglelefteq G\).
4.2. All three Sylow subgroups \(P, Q, R\) are normal in \(G\).
4.3. Since \(|P| = 3\), \(|Q| = 5\), \(|R| = 7\) are pairwise coprime:- \(P \cap (Q R) = \{e\}\) because \(|P| = 3\) and \(|Q R| = 35\) with \(\gcd(3, 35) = 1\).- \(Q \cap R = \{e\}\) because \(\gcd(5, 7) = 1\).
4.4. The product \(P Q R\) is a subgroup of \(G\) with order \(|P Q R| = |P| |Q| |R| = 3 \cdot 5 \cdot 7 = 105 = |G|\), so \(G = P Q R\).
4.5. Therefore \(G\) is isomorphic to the internal direct product:\[G \cong P \times Q \times R \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z}.\]
4.6. By the Chinese Remainder Theorem, \(\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z} \cong \mathbb{Z}/105\mathbb{Z}\), which is cyclic.
5. Conclusion:
$Q$ and $R$ are normal in $G$, $G$ contains a cyclic subgroup of order 35, and normality of $P$ implies $G \cong \mathbb{Z}/105\mathbb{Z}$.