Let G be a group of order 105. Show that G must have a normal Sylow p-subgroup for some prime p dividing 105.

Since the order of the group G is 105, it has prime factorization of the form 3 × 5 × 7. Let n_p denote the number of Sylow p-subgroups of G. By Sylow's theorem, n_p must divide the order of G and be congruent to 1 modulo p for each prime divisor p of the order of G. Therefore, n_3 must be either 1 or 35, n_5 must be either 1 or 21, and n_7 must be either 1 or 15.

Suppose that n_3 = 1. Then there is a unique Sylow 3-subgroup H of G, which is therefore normal in G. In this case, we are done.

Suppose now that n_3 = 35. Then there are 35 Sylow 3-subgroups, and since they all have order 3, they intersect trivially. Therefore, G acts transitively on the set of Sylow 3-subgroups by conjugation. This induces a homomorphism from G into the symmetric group S_35, which has order 35! (35 factorial). However, since the order of G is only 105, this homomorphism must have nontrivial kernel. Let K be the kernel of this homomorphism.

Note that K is a subgroup of G, and its order is divisible by 3, so it must also be divisible by either 5 or 7 (since G has no other prime divisors). Suppose that K has a Sylow q-subgroup P, where q is either 5 or 7. Then P is normal in K, and since K is normal in G, P is also normal in G. In this case, we are done.

Suppose now that K has no Sylow 5-subgroups or Sylow 7-subgroups. Then the order of K is 3^k, where k is at least 2. Since K is a subgroup of G, its order must divide the order of G, which is 3 × 5 × 7 = 105. Therefore, k is at most 3. If k = 3, then K is a Sylow 3-subgroup of G, which contradicts our assumption that n_3 = 35. Therefore, we must have k = 2, and K has order 9.

Let P be a Sylow 5-subgroup of G. Then n_5 divides 21 and is congruent to 1 modulo 5, so n_5 is either 1 or 7. If n_5 = 1, then P is normal in G and we are done. Suppose now that n_5 = 7. Then there are 7 Sylow 5-subgroups, and since they all have order 5, they intersect trivially. Therefore, G acts transitively on the set of Sylow 5-subgroups by conjugation. This induces a homomorphism from G into the symmetric group S_7, which has order 7! (7 factorial). However, since the order of G is only 105, this homomorphism must have nontrivial kernel. Let L be the kernel of this homomorphism.

Note that L is a subgroup of G, and its order is divisible by 5, so it must also be divisible by either 3 or 7 (since G has no other prime divisors). Suppose that L has a Sylow q-subgroup Q, where q is either 3 or 7. Then Q is normal in L, and since L is normal in G, Q is also normal in G. In this case, we are done.

Suppose now that L has no Sylow 3-subgroups or Sylow 7-subgroups. Then the order of L is 5^m, where m is at least 2. Since L is a subgroup of G, its order must divide the order of G, which is 3 × 5 × 7 = 105. Therefore, m is at most 2. If m = 2, then L is a Sylow 5-subgroup of G, which contradicts our assumption that n_5 = 7. Therefore, we must have m = 1, and L has order 5.

Let Q be a Sylow 7-subgroup of G. Then n_7 divides 15 and is congruent to 1 modulo 7, so n_7 is either 1 or 15. If n_7 = 1, then Q is normal in G and we are done. Suppose now that n_7 = 15. Then there are 15 Sylow 7-subgroups, and since they all have order 7, they intersect trivially. Therefore, G acts transitively on the set of Sylow 7-subgroups by conjugation. This induces a homomorphism from G into the symmetric group S_15, which has order 15! (15 factorial). However, since the order of G is only 105, this homomorphism must have nontrivial kernel. Let M be the kernel of this homomorphism.

Note that M is a subgroup of G, and its order is divisible by 7, so it must also be divisible by either 3 or 5 (since G has no other prime divisors). However, M cannot have a Sylow 3-subgroup, since this would imply that n_3 is greater than 1, contradicting our assumption that n_3 is either 1 or 35. Therefore, M must have a Sylow 5-subgroup R, which is normal in M. Since M is normal in G, R is also normal in G. In this case, we are done.

We have now shown that G must have a normal Sylow p-subgroup for some prime p dividing 105. Therefore, the claim is proved.