answer:Since the order of G is 27, we know that |G| = 3^3. According to the Sylow theorems, the number of 3-Sylow subgroups in G, denoted by n_3, must satisfy: 1. n_3 divides the order of G, so n_3 divides 27. 2. n_3 is congruent to 1 modulo 3. The possible values of n_3 are 1 and 3. If n_3 = 1, then there is only one 3-Sylow subgroup, which must be normal in G. If n_3 = 3, then there are three 3-Sylow subgroups, and it is not guaranteed that any of them are normal in G. Let P be a 3-Sylow subgroup of G. Since |P| = 3^2 = 9, P is a group of order 9. The only possible groups of order 9 are C9 (cyclic group of order 9) and C3 × C3 (direct product of two cyclic groups of order 3). However, C3 × C3 has an element of order 3, while every non-identity element of C9 has order 9. Thus, P must be isomorphic to C9. Now, consider the action of G on the set of 3-Sylow subgroups by conjugation. If n_3 = 1, then the single 3-Sylow subgroup is normal in G, and G is a semi-direct product of P with a group of order 3. Since P is abelian, this semi-direct product is actually a direct product, and G is isomorphic to C9 × C3, which is isomorphic to C27 (cyclic group of order 27). If n_3 = 3, then there are three 3-Sylow subgroups, and the action of G on these subgroups is non-trivial. However, this contradicts the fact that P is normal in G, as it is the kernel of the homomorphism from G to the symmetric group S3 induced by the action of G on the 3-Sylow subgroups. Therefore, we must have n_3 = 1, and G is isomorphic to the cyclic group C27.

answer:A group of order 27 has the prime factorization 3^3. According to the Sylow theorems, there exists a Sylow 3-subgroup of order 27. Let's denote this subgroup as P. Case 1: Cyclic group C27 Consider the group Z27, the additive group of integers modulo 27. This group is cyclic, and its order is 27. It has a generator, say 1, such that every element in the group can be written as a power of the generator: {0, 1, 2, ..., 26}. The group operation is addition modulo 27. This group is abelian, and it is isomorphic to the cyclic group C27. Case 2: Non-abelian group of order 27 Consider the group of upper triangular 3x3 matrices with 1's on the diagonal and elements from the field with 3 elements (Z3) in the upper triangle. This group is called the Heisenberg group over Z3, denoted as H(Z3). The group operation is matrix multiplication. The group has 27 elements, as there are 3 choices for each of the 3 upper triangular entries. To show that H(Z3) is non-abelian, consider two matrices A and B: A = | 1 a 1 | | 0 1 b | | 0 0 1 | B = | 1 c 1 | | 0 1 d | | 0 0 1 | Compute the products AB and BA: AB = | 1 a+c 1+a | | 0 1 b+d | | 0 0 1 | BA = | 1 a+c 1+c | | 0 1 b+d | | 0 0 1 | Since AB ≠ BA when a ≠ 0 and c ≠ 0, the group H(Z3) is non-abelian. Conclusion: We have shown that there are at least two distinct groups of order 27: the cyclic group C27 and the non-abelian Heisenberg group H(Z3). Any group of order 27 is isomorphic to either C27 or H(Z3).

answer:To prove that G is isomorphic to the cyclic group C27, we will show that G is generated by a single element, namely a. Since G is a group of order 27, by the Sylow theorems, we know that G is a p-group with p = 3 (since 27 = 3^3). In a p-group, the order of every element divides the order of the group. Thus, the possible orders of elements in G are 1, 3, 9, or 27. We are given that a is an element in G such that a is not the identity element, and the order of a is not 3. Therefore, the order of a must be either 9 or 27. Case 1: The order of a is 9. In this case, the subgroup generated by a, denoted by ⟨a⟩, has 9 elements. Since the order of G is 27, there are 27 - 9 = 18 elements in G that are not in ⟨a⟩. Let b be an element in G that is not in ⟨a⟩. Since the order of every element in G divides the order of G, the order of b must be either 1, 3, 9, or 27. Since b is not in ⟨a⟩, the order of b cannot be 1 or 9. If the order of b is 27, then G is generated by b and is isomorphic to C27, which is what we want to prove. So, let's consider the case where the order of b is 3. Now, consider the set {ab | b is an element in G}. This set has 27 elements since the operation is closed and the elements are distinct (if ab = ac, then b = c by left cancellation). Moreover, for any element x in G, there exists b and c in G such that x = ab and a = ac. Then, x = ab = a(ac) = (aa)c = ac^3, which implies that x is in the set {ab | b is an element in G}. Therefore, the set {ab | b is an element in G} is equal to G. Since the order of a is 9 and the order of b is 3, the order of ab is lcm(9, 3) = 9. Thus, the set {ab | b is an element in G} is a subgroup of G of order 9. However, we have shown that this set is equal to G, which is a contradiction since the order of G is 27. Therefore, the order of b cannot be 3, and we have already shown that if the order of b is 27, then G is isomorphic to C27. Case 2: The order of a is 27. In this case, G is generated by a, and G is isomorphic to the cyclic group C27. In both cases, we have shown that G is isomorphic to the cyclic group C27.

answer:To determine the number of non-isomorphic groups of order 27, we can use the fact that 27 = 3^3. According to the Sylow theorems, the number of groups of order p^n (where p is a prime number and n is a positive integer) is determined by the number of possible factorizations of the group's order. In this case, we have the following factorizations of 27: 1. 27 = 3^3 Since 27 is a prime power, we can use the following theorem: Theorem: If G is a group of order p^n, where p is a prime number and n is a positive integer, then G has a normal subgroup of order p^i for every 0 ≤ i ≤ n. Using this theorem, we can find the number of non-isomorphic groups of order 27: 1. For the factorization 27 = 3^3, we have a group of order 3^3. By the theorem, this group has normal subgroups of order 3^0 = 1, 3^1 = 3, and 3^2 = 9. There are two possibilities for the group structure: either it is a direct product of its subgroups, or it is a semi-direct product of its subgroups. In the first case, the group is the direct product of a group of order 3 and a group of order 9. Since there is only one group of order 3 (up to isomorphism), namely the cyclic group C3, and there is only one group of order 9 (up to isomorphism), namely the cyclic group C9, the direct product is C3 x C9, which is isomorphic to the cyclic group C27. In the second case, the group is a semi-direct product of a group of order 3 and a group of order 9. There are two possible semi-direct products, depending on the action of the group of order 3 on the group of order 9. One of these actions is trivial, which gives the direct product C3 x C9 (which we already counted). The other action is non-trivial, which gives a non-abelian group of order 27. Thus, there are two non-isomorphic groups of order 27: the cyclic group C27 and a non-abelian group.