Here is a challenging math problem from the field of Abstract Algebra:
Let G be a group of order 105. Show that G must have a normal Sylow p-subgroup for some prime p dividing 105.
Solution:
First, note that the prime factorization of 105 is 3 x 5 x 7. By the Sylow theorems, G must have Sylow p-subgroups for each of these primes. Let P3, P5, and P7 denote the Sylow 3-subgroup, Sylow 5-subgroup, and Sylow 7-subgroup of G, respectively.
Since the order of each Sylow p-subgroup is a power of p and since the only possible orders for a Sylow 3-subgroup, Sylow 5-subgroup, or Sylow 7-subgroup are 3, 5, and 7, respectively, the orders of these Sylow subgroups must be 3, 5, and 7, respectively.
Let n3, n5, and n7 denote the number of Sylow 3-subgroups, Sylow 5-subgroups, and Sylow 7-subgroups of G, respectively. By the Sylow theorems, we have:
n3 ≡ 1 (mod 3) and n3 | 35
n5 ≡ 1 (mod 5) and n5 | 21
n7 ≡ 1 (mod 7) and n7 | 15
From these congruences, we see that n3 = 1 or 7, n5 = 1 or 21, and n7 = 1 or 15.
If n7 = 1, then P7 is normal in G, since any Sylow 7-subgroup is conjugate to P7. If n7 = 15, then G acts transitively on the set of Sylow 7-subgroups by conjugation, and the stabilizer of P7 is a subgroup of order 21. By the Sylow theorems, there must be a Sylow 3-subgroup in this subgroup of order 21, which is then normal in G.
If n5 = 1, then P5 is normal in G, since any Sylow 5-subgroup is conjugate to P5. If n5 = 21, then G acts transitively on the set of Sylow 5-subgroups by conjugation, and the stabilizer of P5 is a subgroup of order 3 x 7 = 21. By the Sylow theorems, there must be a Sylow 7-subgroup in this subgroup of order 21, which is then normal in G.
If n3 = 1, then P3 is normal in G, since any Sylow 3-subgroup is conjugate to P3. If n3 = 7, then G acts transitively on the set of Sylow 3-subgroups by conjugation, and the stabilizer of P3 is a subgroup of order 5 x 7 = 35. By the Sylow theorems, there must be a Sylow 5-subgroup in this subgroup of order 35, which is then normal in G.
Therefore, in all cases, 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.
Answers & Comments
Verified answer
Here is a challenging math problem from the field of Abstract Algebra:
Let G be a group of order 105. Show that G must have a normal Sylow p-subgroup for some prime p dividing 105.
Solution:
First, note that the prime factorization of 105 is 3 x 5 x 7. By the Sylow theorems, G must have Sylow p-subgroups for each of these primes. Let P3, P5, and P7 denote the Sylow 3-subgroup, Sylow 5-subgroup, and Sylow 7-subgroup of G, respectively.
Since the order of each Sylow p-subgroup is a power of p and since the only possible orders for a Sylow 3-subgroup, Sylow 5-subgroup, or Sylow 7-subgroup are 3, 5, and 7, respectively, the orders of these Sylow subgroups must be 3, 5, and 7, respectively.
Let n3, n5, and n7 denote the number of Sylow 3-subgroups, Sylow 5-subgroups, and Sylow 7-subgroups of G, respectively. By the Sylow theorems, we have:
n3 ≡ 1 (mod 3) and n3 | 35
n5 ≡ 1 (mod 5) and n5 | 21
n7 ≡ 1 (mod 7) and n7 | 15
From these congruences, we see that n3 = 1 or 7, n5 = 1 or 21, and n7 = 1 or 15.
If n7 = 1, then P7 is normal in G, since any Sylow 7-subgroup is conjugate to P7. If n7 = 15, then G acts transitively on the set of Sylow 7-subgroups by conjugation, and the stabilizer of P7 is a subgroup of order 21. By the Sylow theorems, there must be a Sylow 3-subgroup in this subgroup of order 21, which is then normal in G.
If n5 = 1, then P5 is normal in G, since any Sylow 5-subgroup is conjugate to P5. If n5 = 21, then G acts transitively on the set of Sylow 5-subgroups by conjugation, and the stabilizer of P5 is a subgroup of order 3 x 7 = 21. By the Sylow theorems, there must be a Sylow 7-subgroup in this subgroup of order 21, which is then normal in G.
If n3 = 1, then P3 is normal in G, since any Sylow 3-subgroup is conjugate to P3. If n3 = 7, then G acts transitively on the set of Sylow 3-subgroups by conjugation, and the stabilizer of P3 is a subgroup of order 5 x 7 = 35. By the Sylow theorems, there must be a Sylow 5-subgroup in this subgroup of order 35, which is then normal in G.
Therefore, in all cases, 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.