answer:Let G be a group of order 36. Since 36 = 2^2 * 3^2, by the Sylow theorems, we know that there are Sylow 2-subgroups and Sylow 3-subgroups in G. Let's find the possible number of these subgroups and their structures. For Sylow 2-subgroups: By the Sylow theorems, the number of Sylow 2-subgroups, n_2, must divide 3^2 = 9 and n_2 ≡ 1 (mod 2). So, n_2 can be 1 or 9. If n_2 = 1, then there is a unique Sylow 2-subgroup, which is normal. If n_2 = 9, then there are 9 Sylow 2-subgroups, and they are not normal. The possible structures for a group of order 4 are C4 (cyclic) and V4 (Klein-4 group). For Sylow 3-subgroups: By the Sylow theorems, the number of Sylow 3-subgroups, n_3, must divide 2^2 = 4 and n_3 ≡ 1 (mod 3). So, n_3 can be 1 or 4. If n_3 = 1, then there is a unique Sylow 3-subgroup, which is normal. If n_3 = 4, then there are 4 Sylow 3-subgroups, and they are not normal. The possible structures for a group of order 9 are C9 (cyclic) and C3 × C3. Now, let's analyze the possible structures for G: 1. If G has a unique normal Sylow 2-subgroup and a unique normal Sylow 3-subgroup, then G is isomorphic to C4 × C9 or C4 × (C3 × C3) or V4 × C9 or V4 × (C3 × C3). The first two cases are abelian, and the last two cases are non-abelian. 2. If G has a unique normal Sylow 2-subgroup and 4 non-normal Sylow 3-subgroups, then G is a semi-direct product of C4 or V4 with C3 × C3. This gives four non-abelian groups. 3. If G has 9 non-normal Sylow 2-subgroups and a unique normal Sylow 3-subgroup, then G is a semi-direct product of C9 or (C3 × C3) with C4. This gives two non-abelian groups. 4. If G has 9 non-normal Sylow 2-subgroups and 4 non-normal Sylow 3-subgroups, then G is isomorphic to the alternating group A4 × C3, which is non-abelian. In total, we have 2 abelian groups (C36 and C4 × C9) and 9 non-abelian groups. Therefore, every group of order 36 is isomorphic to either the cyclic group C36, the alternating group A4 × C3, or one of the nine non-abelian groups of order 36.

answer:To determine the number of non-isomorphic groups of order 36, we will 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 36, which can be factored as 2^2 * 3^2. Thus, we need to find the Sylow 2-subgroups and Sylow 3-subgroups of G. 1. Sylow 2-subgroups: The number of Sylow 2-subgroups, denoted by n_2, must divide 3^2 = 9 and n_2 ≡ 1 (mod 2). The possible values for n_2 are 1 and 9. If n_2 = 1, then there is a unique Sylow 2-subgroup of order 4, which must be isomorphic to either Z_4 or Z_2 × Z_2. If n_2 = 9, then all Sylow 2-subgroups are isomorphic to Z_2 × Z_2. 2. Sylow 3-subgroups: The number of Sylow 3-subgroups, denoted by n_3, must divide 2^2 = 4 and n_3 ≡ 1 (mod 3). The possible values for n_3 are 1 and 4. If n_3 = 1, then there is a unique Sylow 3-subgroup of order 9, which must be isomorphic to either Z_9 or Z_3 × Z_3. If n_3 = 4, then all Sylow 3-subgroups are isomorphic to Z_3 × Z_3. Now, we can list the possible isomorphism classes of groups of order 36 based on the Sylow subgroups: 1. If G has a unique Sylow 2-subgroup isomorphic to Z_4 and a unique Sylow 3-subgroup isomorphic to Z_9, then G is isomorphic to Z_4 × Z_9, which is isomorphic to Z_36. 2. If G has a unique Sylow 2-subgroup isomorphic to Z_4 and a unique Sylow 3-subgroup isomorphic to Z_3 × Z_3, then G is isomorphic to Z_4 × (Z_3 × Z_3), which is isomorphic to Z_4 × Z_3 × Z_3. 3. If G has a unique Sylow 2-subgroup isomorphic to Z_2 × Z_2 and a unique Sylow 3-subgroup isomorphic to Z_9, then G is isomorphic to (Z_2 × Z_2) × Z_9, which is isomorphic to Z_2 × Z_2 × Z_9. 4. If G has 9 Sylow 2-subgroups isomorphic to Z_2 × Z_2 and 4 Sylow 3-subgroups isomorphic to Z_3 × Z_3, then G is a semi-direct product of Z_2 × Z_2 and Z_3 × Z_3. There are two non-isomorphic semi-direct products in this case, which we denote by (Z_2 × Z_2) ⋊ (Z_3 × Z_3)_1 and (Z_2 × Z_2) ⋊ (Z_3 × Z_3)_2. In total, there are 5 non-isomorphic groups of order 36: 1. Z_36 2. Z_4 × Z_3 × Z_3 3. Z_2 × Z_2 × Z_9 4. (Z_2 × Z_2) ⋊ (Z_3 × Z_3)_1 5. (Z_2 × Z_2) ⋊ (Z_3 × Z_3)_2

answer:There are four non-isomorphic groups of order 36. These groups are: 1. Cyclic group C_36 2. Direct product of two cyclic groups C_9 × C_4 3. Direct product of two cyclic groups C_3 × C_12 4. The semi-direct product of C_9 and C_4, denoted as C_9 ⋊ C_4 These groups are non-isomorphic because they have different structures and cannot be mapped onto one another by an isomorphism.

answer:To determine which non-abelian group of order 36 the given group is isomorphic to, we first need to analyze the structure of the group G. 1. The order of G is 36, which has prime factorization 2^2 * 3^2. 2. The group G is non-abelian, which means that the group operation is not commutative. 3. The multiplication table provided shows that the group operation is closed and has an identity element e. Now, we know that there are two non-abelian groups of order 36 up to isomorphism: the dihedral group D_18 and the direct product of the symmetric group S_3 and the cyclic group C_6, denoted as S_3 × C_6. To determine which group G is isomorphic to, we can look for properties that distinguish these two groups: 1. D_18 has 18 elements of order 2, while S_3 × C_6 has 6 elements of order 2. 2. D_18 has a cyclic subgroup of order 18, while S_3 × C_6 does not. Let's analyze the given multiplication table to see which properties hold for G: 1. Elements of order 2: We can see that there are 6 elements of order 2 in G (b, d, h, o, r, and s). This matches the property of S_3 × C_6. 2. Cyclic subgroup of order 18: There is no cyclic subgroup of order 18 in G, as no element generates a subgroup of order 18. This also matches the property of S_3 × C_6. Based on these properties, we can conclude that the given group G is isomorphic to the non-abelian group S_3 × C_6.