Question

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

Answer 1

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.

Answer 2

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.

Answer 3

To show that a group G of order 105 must have a normal Sylow p-subgroup for some prime p dividing 105, we'll utilize the Sylow theorems.

The Sylow theorems provide important insights into the structure of groups by studying the properties of their p-subgroups, where p is a prime number. Let's consider the prime factorization of 105:

105 = 3 × 5 × 7.

According to the Sylow theorems:

1. The number of Sylow p-subgroups of G, denoted by n_p, must satisfy the following conditions: a) n_p ≡ 1 (mod p), which means that n_p leaves a remainder of 1 when divided by p. b) n_p divides the order of G, in this case, 105.

2. Any two Sylow p-subgroups are conjugate to each other, meaning that for any two Sylow p-subgroups P and Q, there exists an element g in G such that Q = gP(g^(-1)).

3. The number of Sylow p-subgroups, n_p, is equal to the index of the normalizer of any Sylow p-subgroup in G.

Now, let's consider the possible Sylow p-subgroups for the prime factors of 105:

For p = 3: The number of Sylow 3-subgroups, n_3, must divide 105 and leave a remainder of 1 when divided by 3. The possible values for n_3 are 1, 7, 35, and 105. However, since 7 and 35 do not satisfy the condition n_3 ≡ 1 (mod 3), they can be ruled out. Therefore, n_3 = 1 or 105.

Similarly, for p = 5 and p = 7: The number of Sylow 5-subgroups, n_5, must divide 105 and leave a remainder of 1 when divided by 5. The possible values for n_5 are 1, 21, and 105. The number of Sylow 7-subgroups, n_7, must divide 105 and leave a remainder of 1 when divided by 7. The possible values for n_7 are 1, 15, and 105.

Since 105 is not a prime power, we know that n_p ≠ 105 for any prime p. Therefore, the only possible values for n_3, n_5, and n_7 are 1.

Now, let's suppose there are n_3 = n_5 = n_7 = 1 Sylow subgroups for the primes 3, 5, and 7, respectively. Denote these subgroups as P_3, P_5, and P_7.

Since there is only one Sylow p-subgroup for each prime, they are all unique. Thus, the intersection of any two of these subgroups must be the identity element, i.e., P_3 ∩ P_5 = {e}, P_3 ∩ P_7 = {e}, and P_5 ∩ P_7 = {e}.

Consider the product H = P_3 × P_5 × P_7, which is a subgroup of G since each P_i is a subgroup of G. By construction, the order of H is 3 × 5 × 7 = 105, which means H is a subgroup of G with the same order as G.

Since the intersection of any two Sylow subgroups is trivial, H is is