answer:To determine if G is isomorphic to any of the given groups, we need to analyze its Sylow subgroups. The order of a group G is 72, which can be factored as 2^3 * 3^2. According to Sylow's Theorems, the number of Sylow 2-subgroups (n_2) and Sylow 3-subgroups (n_3) must satisfy: 1. n_2 divides 3^2 = 9, and n_2 is congruent to 1 mod 2. So, n_2 can be 1, 3, or 9. 2. n_3 divides 2^3 = 8, and n_3 is congruent to 1 mod 3. So, n_3 can be 1, 4, or 8. Now, let's examine the possibilities for G: 1. If G is isomorphic to the cyclic group C72, then G must have a unique Sylow 2-subgroup and a unique Sylow 3-subgroup (since a cyclic group has a unique subgroup of each order dividing the group order). Thus, n_2 = n_3 = 1. So, G is isomorphic to C72 if and only if it has a unique Sylow 2-subgroup and a unique Sylow 3-subgroup. 2. If G is isomorphic to the alternating group A4 × C3 × C2, then G has normal Sylow 3-subgroups (since A4 has normal Sylow 3-subgroups). In this case, n_3 = 1 or 4. If n_3 = 4, then n_2 must also be 9 (since the Sylow theorems require n_2 and n_3 to be relatively prime). So, G is isomorphic to A4 × C3 × C2 if and only if it has normal Sylow 3-subgroups and 9 Sylow 2-subgroups. 3. If G is isomorphic to one of the 51 non-abelian groups of order 72, then G cannot be isomorphic to C72 or A4 × C3 × C2. So, G must have at least two distinct Sylow 2-subgroups and at least two distinct Sylow 3-subgroups. In, To determine whether a group G of order 72 is isomorphic to the cyclic group C72, the alternating group A4 × C3 × C2, or one of the 51 non-abelian groups of order 72, we need more information about the group G. Specifically, we need to know about its structure and properties, such as whether it is abelian or not, its subgroups, and its elements' orders. Here are some steps to help you determine the isomorphism: 1. Check if G is abelian: If G is abelian, then it cannot be isomorphic to any non-abelian group of order 72. In this case, you can compute the number of elements of each order in G and compare it with the number of elements of each order in C72. If they match, then G is isomorphic to C72. 2. Check if G has a normal subgroup H with |H| = 24 and G/H is cyclic: If G has such a normal subgroup, then by the Fundamental Theorem of Homomorphisms, G is isomorphic to a semidirect product H ⋊ K, where K is a subgroup of G/H with |K| = 3. Since G/H is cyclic, K must be cyclic too. Therefore, G is isomorphic to A4 × C3 × C2. 3. If G is not abelian and doesn't have the required normal subgroup, then compare its properties with the known non-abelian groups of order 72: There are 51 such groups, and they have various properties that differ from each other. You would need to analyze the structure of G and find a matching non-abelian group among the 51 possibilities. This might involve finding specific subgroups, checking for automorphisms, or using other group theory techniques. Without additional information about the group G, it is impossible to provide a definitive answer regarding its isomorphism to any of the mentioned groups. However, these steps should help you determine the isomorphism when given enough details about G.,To determine whether G is isomorphic to one of these groups, we can first analyze the structure of G using the Sylow theorems. The order of G is 72, which can be factored as 2^3 * 3^2. According to the Sylow theorems, the number of Sylow 2-subgroups (n_2) must divide 3^2 and be congruent to 1 mod 2. The possible values for n_2 are 1 and 9. Similarly, the number of Sylow 3-subgroups (n_3) must divide 2^3 and be congruent to 1 mod 3. The possible values for n_3 are 1 and 4. Case 1: n_2 = 1 and n_3 = 1 If both n_2 and n_3 are 1, then G has a unique Sylow 2-subgroup and a unique Sylow 3-subgroup, both of which are normal. In this case, G is the direct product of these two normal subgroups, and G is isomorphic to the cyclic group C72. Case 2: n_2 = 1 and n_3 = 4 In this case, G has a unique normal Sylow 2-subgroup of order 8 and four Sylow 3-subgroups of order 9. The group G cannot be isomorphic to A4 × C3 × C2 in this case, as A4 × C3 × C2 has a unique Sylow 3-subgroup. Therefore, G must be isomorphic to one of the non-abelian groups of order 72. Case 3: n_2 = 9 and n_3 = 1 In this case, G has nine Sylow 2-subgroups of order 8 and a unique normal Sylow 3-subgroup of order 9. The group G cannot be isomorphic to A4 × C3 × C2 in this case, as A4 × C3 × C2 has a unique Sylow 2-subgroup. Therefore, G must be isomorphic to one of the non-abelian groups of order 72. Case 4: n_2 = 9 and n_3 = 4 In this case, G has nine Sylow 2-subgroups of order 8 and four Sylow 3-subgroups of order 9. The group G cannot be isomorphic to A4 × C3 × C2 in this case, as A4 × C3 × C2 has a unique Sylow 2-subgroup and a unique Sylow 3-subgroup. Therefore, G must be isomorphic to one of the non-abelian groups of order 72. In conclusion, without more information about the group G, we cannot determine a specific isomorphism. However, we have shown that G can be isomorphic to the cyclic group C72 or one of the 51 non-abelian groups of order 72, but not to the alternating group A4 × C3 × C2.

answer:To determine the number of non-abelian groups of order 72, we first need to find the prime factorization of 72. The prime factorization of 72 is 2^3 * 3^2. Now, we will use the formula for the number of groups of a given order, which is the product of the p-group counts for each prime factor. The p-group counts are given by the partitions of the exponents in the prime factorization. For 2^3, there are three partitions: (3), (2, 1), and (1, 1, 1). These correspond to the following cyclic groups: C_8, C_4 × C_2, and C_2 × C_2 × C_2. For 3^2, there are two partitions: (2) and (1, 1). These correspond to the cyclic groups: C_9 and C_3 × C_3. Now, we will find the direct products of these groups: 1. C_8 × C_9 2. C_4 × C_2 × C_9 3. C_2 × C_2 × C_2 × C_9 4. C_8 × C_3 × C_3 5. C_4 × C_2 × C_3 × C_3 6. C_2 × C_2 × C_2 × C_3 × C_3 Out of these six possibilities, the first one, C_8 × C_9, is abelian, so we can exclude it. The remaining five possibilities are non-abelian groups of order 72: 1. C_4 × C_2 × C_9 2. C_2 × C_2 × C_2 × C_9 3. C_8 × C_3 × C_3 4. C_4 × C_2 × C_3 × C_3 5. C_2 × C_2 × C_2 × C_3 × C_3 So, there are 5 non-abelian groups of order 72.,There are 19 non-abelian groups of order 72. However, it is not feasible to name all 19 groups individually here. There are various ways to classify and describe these groups, such as using their names or their presentation in terms of generators and relations, but it would be too large of an undertaking to list them all. It is recommended to consult a reference book on group theory or use computational tools for a more detailed study of these groups., There are 14 non-isomorphic non-abelian groups of order 72 up to isomorphism. It is difficult to name each one explicitly, but I can give you a brief description of how they are constructed. 1. SmallGroup(72, i) for i = 1, 2, ..., 14 in the GAP system's Small Group Library (https://groupprops.subwiki.org/wiki/Groups_of_order_72). 2. (C2 x C2 x C2 x C3) ⋊ C6 ( semidirect product) 3. (C2 x C2 x C6) ⋊ C3 ( semidirect product) 4. (C2 x C2 x C3) ⋊ Q8 ( semidirect product where Q8 is the quaternion group of order 8) 5. (C2 x C2 x C3) ⋊ D8 ( semidirect product where D8 is the dihedral group of order 8) 6. (C2 x C2 x C3) ⋊ SD16 ( semidirect product where SD16 is the semidihedral group of order 16) 7. (C2 x C2 x C3) ⋊ QD16 ( semidirect product where QD16 is the generalized quaternion group of order 16) 8. (C2 x C2 x C3) ⋊ C8 ( semidirect product where C8 is the cyclic group of order 8) 9. (C2 x C2 x C6) ⋊ Q8 ( semidirect product) 10. (C2 x C2 x C6) ⋊ D8 ( semidirect product) 11. (C2 x C2 x C6) ⋊ SD16 ( semidirect product) 12. (C2 x C2 x C6) ⋊ QD16 ( semidirect product) 13. (C2 x C2 x C6) ⋊ C8 ( semidirect product) 14. (C3 x C3 x C3) ⋊ Q8 ( semidirect product) Please note that constructing and checking whether these groups are non-isomorphic would require a computational tool like GAP or MAGMA.

answer:To find the number of non-isomorphic groups of order 72, we can use Sylow's theorems. First, let's find the Sylow 2-subgroups and Sylow 3-subgroups. The order of a Sylow 2-subgroup is 2^k where 2^k mid 72 and 2^{k+1} nmid 72, so k=3. Thus, there are n_2 equiv 1 pmod 2 and n_2 mid 9 Sylow 2-subgroups. Since n_2 cannot be 1 (as that would mean the group is Abelian, but it's not), n_2=3, 7. But |N_G(P_2)|=8m, where P_2 is a Sylow 2-subgroup, so |N_G(P_2)| cannot divide 72. Therefore, n_2=3, which means there are three Sylow 2-subgroups. Similarly, for the Sylow 3-subgroups, the order is 3^l where 3^l mid 72 and 3^{l+1} nmid 72, so l=1. Thus, there are n_3 equiv 1 pmod 3 and n_3 mid 8 Sylow 3-subgroups. As n_3 cannot be 1, n_3=4. Now, let's analyze the possible cases: 1) G has a normal Sylow 2-subgroup H and a normal Sylow 3-subgroup K. In this case, H cong mathbb{Z}_8 or Q_8 (the quaternion group), and K cong mathbb{Z}_9. Since both subgroups are Abelian, G is also Abelian, so G cong mathbb{Z}_8 times mathbb{Z}_9. 2) G has a normal Sylow 2-subgroup H and n_3 = 4. In this case, H cong mathbb{Z}_8 or Q_8, and we have four Sylow 3-subgroups that form a group isomorphic to mathbb{Z}_3 times mathbb{Z}_3. This group acts via conjugation on the Sylow 2-subgroups. As |H| = 8, the only possible action is the trivial action, meaning that these are all normal subgroups of G. Thus, G has a subgroup isomorphic to mathbb{Z}_8 times (mathbb{Z}_3 times mathbb{Z}_3). However, as this subgroup has index 2, it must be a normal subgroup itself, so G is isomorphic to the direct product mathbb{Z}_8 times (mathbb{Z}_3 times mathbb{Z}_3). 3) G has n_2 = 3, and n_3 = 4. In this case, neither Sylow 2-subgroups nor Sylow 3-subgroups are normal. Since they are all conjugate, we can compute the number of isomorphism types by finding the number of ways to choose a Sylow 2-subgroup multiplied by the number of ways to choose a Sylow 3-subgroup. However, since n_2 = 3, we have only one isomorphism type. So, we have found two non-isomorphic groups of order 72: one Abelian group mathbb{Z}_8 times mathbb{Z}_9 and one non-Abelian group isomorphic to the semi-direct product mathbb{Z}_8 rtimes (mathbb{Z}_3 times mathbb{Z}_3). However, this isn't the end of the story. We have only considered groups that are an extension of mathbb{Z}_9 by mathbb{Z}_8 or an extension of mathbb{Z}_3 times mathbb{Z}_3 by mathbb{Z}_8. But there are more possibilities. To find all non-isomorphic groups, one can either use the O'Nan-Scott theorem or compute the Schur multiplier and outer automorphism group of the Sylow 2-subgroups. It turns out that there are, in total, five non-isomorphic groups of order 72: 1. The Abelian group mathbb{Z}_8 times mathbb{Z}_9. 2. The group mathbb{Z}_8 rtimes (mathbb{Z}_3 times mathbb{Z}_3). 3. The generalized dihedral group Dic_{16}, which is an extension of mathbb{Z}_8 by mathbb{Z}_2, where mathbb{Z}_2 acts as the inverse map. 4. The non-Abelian group M_{16} of order 16, which is the only non-Abelian group with 16 elements. 5. The non-Abelian group SD_{16}, which is a semi-direct product mathbb{Z}_8 rtimes mathbb{Z}_2, where mathbb{Z}_2 acts via the automorphism x to x^3 of mathbb{Z}_8. The last three groups only have elements of orders 1, 2, 4, and 8. Therefore, these are not considered in the original problem, which only asks for groups isomorphic to the semi-direct product mathbb{Z}_8 rtimes (mathbb{Z}_3 times mathbb{Z}_3). However, rigorously speaking, there are five non-isomorphic groups of order 72.,To find the number of non-isomorphic groups of order 72, we can use the following steps: 1. Prime factorize the order: 72 = 2^3 * 3^2 2. Determine the possible partitions of the exponents: (3, 2) 3. Calculate the number of non-isomorphic groups for each partition using the formula n(G) = (p + e - 1)C(e-1), where p is the number of prime factors, e is the exponent of the highest power of the prime factors, and C denotes the binomial coefficient. For (3, 2), we have n(G) = (2 + 3 - 1)C(3 - 1) = 4C2 = 6. Thus, there are 6 non-isomorphic groups of order 72. The names of these groups are as follows: 1. The cyclic group of order 72: C_72 2. The direct product of two cyclic groups: C_9 × C_8 3. The direct product of a cyclic group and a non-abelian group of order 8: C_9 × D_4 4. The direct product of a cyclic group and a non-abelian group of order 9: C_8 × Q_8 5. The semidirect product of a cyclic group and a non-abelian group of order 8: C_9 ⋊ D_4 6. The semidirect product of a cyclic group and a non-abelian group of order 9: C_8 ⋊ Q_8 These are the six non-isomorphic groups of order 72.,To find the number of non-isomorphic groups of order 72, we can use the prime factorization of 72 and Sylow theorems. The prime factorization of 72 is 2^3 * 3^2. Now, we will consider the possible Sylow subgroups for each prime factor. For the 2-Sylow subgroups, there are three possibilities: C_8, C_4 x C_2, and C_2 x C_2 x C_2. For the 3-Sylow subgroups, there are two possibilities: C_9 and C_3 x C_3. Now, we can combine these possibilities to form different groups: 1. C_8 x C_9: This is a direct product of two cyclic groups, which results in a cyclic group of order 72, denoted as C_72. 2. C_8 x (C_3 x C_3): This is a direct product of C_8 and C_3 x C_3, which results in a group of order 72, denoted as C_8 x C_3 x C_3. 3. (C_4 x C_2) x C_9: This is a direct product of C_4 x C_2 and C_9, which results in a group of order 72, denoted as C_4 x C_2 x C_9. 4. (C_4 x C_2) x (C_3 x C_3): This is a direct product of C_4 x C_2 and C_3 x C_3, which results in a group of order 72, denoted as C_4 x C_2 x C_3 x C_3. 5. (C_2 x C_2 x C_2) x C_9: This is a direct product of C_2 x C_2 x C_2 and C_9, which results in a group of order 72, denoted as C_2 x C_2 x C_2 x C_9. 6. (C_2 x C_2 x C_2) x (C_3 x C_3): This is a direct product of C_2 x C_2 x C_2 and C_3 x C_3, which results in a group of order 72, denoted as C_2 x C_2 x C_2 x C_3 x C_3. However, some of these groups are isomorphic to each other. After analyzing the groups, we find that there are 5 non-isomorphic groups of order 72: 1. C_72: The cyclic group of order 72. 2. C_8 x C_3 x C_3: The direct product of C_8 and C_3 x C_3. 3. C_4 x C_2 x C_9: The direct product of C_4 x C_2 and C_9. 4. C_4 x C_2 x C_3 x C_3: The direct product of C_4 x C_2 and C_3 x C_3. 5. C_2 x C_2 x C_2 x C_9: The direct product of C_2 x C_2 x C_2 and C_9. So, there are 5 non-isomorphic groups of order 72.

answer:To find the possible number of non-isomorphic groups of order 72, we can use the fact that the order of a group is the product of the orders of its Sylow subgroups. The order of a group G is 72, which can be factored as 2^3 * 3^2. Thus, we need to find the Sylow subgroups of G. For the 2-Sylow subgroups, by Sylow's Theorem, the number of 2-Sylow subgroups, n_2, must divide 3^2 = 9, and n_2 ≡ 1 (mod 2). The possible values for n_2 are 1 and 9. For the 3-Sylow subgroups, the number of 3-Sylow subgroups, n_3, must divide 2^3 = 8, and n_3 ≡ 1 (mod 3). The possible values for n_3 are 1 and 4. Now, we can list the possible combinations of Sylow subgroups and their orders: 1. n_2 = 1, n_3 = 1: There is only one 2-Sylow subgroup of order 8 and one 3-Sylow subgroup of order 9. The group is a direct product of these subgroups, and is isomorphic to Z_8 x Z_9, which is abelian. 2. n_2 = 1, n_3 = 4: There is only one 2-Sylow subgroup of order 8 and four 3-Sylow subgroups of order 9. The group is a semi-direct product of these subgroups, and is isomorphic to Z_8 x Z_9 ⋊ Z_2, which is non-abelian. 3. n_2 = 9, n_3 = 1: There are nine 2-Sylow subgroups of order 8 and one 3-Sylow subgroup of order 9. The group is a semi-direct product of these subgroups, and is isomorphic to (Z_8 x Z_9) ⋊ Z_3, which is non-abelian. 4. n_2 = 9, n_3 = 4: There are nine 2-Sylow subgroups of order 8 and four 3-Sylow subgroups of order 9. The group is a semi-direct product of these subgroups, and is isomorphic to (Z_8 x Z_9) ⋊ (Z_2 x Z_2), which is non-abelian. In total, there are 4 non-isomorphic groups of order 72, and 3 of them are non-abelian. List of all non-isomorphic groups of order 72: 1. Z_8 x Z_9 (abelian) 2. Z_8 x Z_9 ⋊ Z_2 (non-abelian) 3. (Z_8 x Z_9) ⋊ Z_3 (non-abelian) 4. (Z_8 x Z_9) ⋊ (Z_2 x Z_2) (non-abelian),The order of a group is the number of its elements. A group of order 72 has 72 elements. There are many methods to determine the number of non-isomorphic groups of order 72, but one of the most straightforward ways is to use the Sylow theorems. In this case, we can proceed given the prime factorization of 72, i.e., 72 = 2^3 * 3^2. The Sylow theorems allow us to determine the number of Sylow p-subgroups for each prime factor of the group order. For the prime factorization 2^3 * 3^2, we have: - The number of Sylow 2-subgroups is either 1 or 3 or 9, but not greater than 3 (otherwise, we would have more than 72 elements in the group). - The number of Sylow 3-subgroups is either 1 or 4. Now, the possible number of non-isomorphic groups of order 72 can be found by considering all possible combinations of Sylow subgroups. We can break down this into a few cases: 1. Exactly one Sylow 2-subgroup and one Sylow 3-subgroup. 2. Exactly one Sylow 2-subgroup and four Sylow 3-subgroups. 3. Three Sylow 2-subgroups and one Sylow 3-subgroup. 4. Nine Sylow 2-subgroups and one Sylow 3-subgroup. For each of these cases, we must examine how the Sylow subgroups interact to form the group. After considering these cases, we find that there are 13 non-isomorphic groups of order 72. Some of these groups are abelian, and some are non-abelian. Out of the 13 groups, 11 of them are non-abelian, and the remaining 2 are abelian. We cannot list the specific non-isomorphic groups of order 72 here since the number of possible groups is quite large, and listing them all would be impractical., To find the number of non-isomorphic groups of order 72, we can use the Sylow Theorems. First, let's find the possible numbers of Sylow 3-subgroups (since 3 is a prime factor of 72 and 3^3 > 72). The number of Sylow 3-subgroups, denoted by n_3, divides the order of the group and is congruent to 1 modulo 3. So, the possible values for n_3 are 1 and 4. If n_3 = 1, then there is a unique Sylow 3-subgroup, which must be normal. In this case, the Sylow 3-subgroup intersects trivially with its conjugates, and thus it has 27 elements. The quotient group has order 72/3 = 24. Since 24 has only one Sylow 2-subgroup (which is normal), the group is isomorphic to Z_3 ⋊ Z_8, where Z_n denotes the cyclic group of order n. If n_3 = 4, then there are 4 Sylow 3-subgroups, and they form a single orbit under conjugation. The action of the group on this orbit gives a homomorphism from the group to S_4, the symmetric group on 4 letters. The kernel of this homomorphism has index at most 24 (the order of S_4), so it is either the whole group (in which case the group is isomorphic to S_4) or a subgroup of order 3 (in which case the group is isomorphic to A_4, the alternating group on 4 letters). We have found three possible non-isomorphic groups of order 72: Z_3 ⋊ Z_8, S_4, and A_4. To determine which ones are abelian, we note that Z_3 ⋊ Z_8 is not abelian because Z_8 is not abelian (it has an element of order 4). Also, both S_4 and A_4 are non-abelian because they contain the non-abelian subgroup S_3. Therefore, there is one abelian group of order 72 (Z_3 ⋊ Z_8) and two non-abelian groups of order 72 (S_4 and A_4). To summarize: Number of non-isomorphic groups of order 72: 3 Number of non-abelian groups of order 72: 2 List of non-isomorphic groups of order 72: Z_3 ⋊ Z_8, S_4, A_4