answer:To prove that G is isomorphic to the dihedral group D3, we will first show that G has a cyclic subgroup of order 3, and then we will show that G can be generated by two elements, one of order 2 and the other of order 3. Let G be a group of order 6 with exactly 3 distinct elements of order 2. By the Cauchy's theorem, there exists an element of order 3 in G. Let x be such an element. Then, the cyclic subgroup generated by x, denoted by ⟨x⟩, has order 3. Since the order of G is 6, there are 6 - 3 = 3 elements outside of ⟨x⟩. Let these elements be y, z, and w. Now, we know that the order of y, z, and w is 2, as there are exactly 3 distinct elements of order 2 in G. Thus, y^2 = z^2 = w^2 = e, where e is the identity element of G. We will now show that G can be generated by x and y. Since x has order 3 and y has order 2, their product xy has order 6 or 2. If the order of xy is 6, then G = ⟨x, y⟩, and we are done. If the order of xy is 2, then (xy)^2 = x^2y^2 = e. Since x^2 ≠ e and y^2 = e, we have x^2 = y^(-1)xy. Thus, x^2 is a product of two elements of order 2, which implies that x^2 = z or x^2 = w (since x^2 ≠ y^2 = e). Without loss of generality, let x^2 = z. Then, G = {e, x, x^2, y, yx, yx^2} = {e, x, z, y, yx, yz}. Since x has order 3 and y has order 2, G is generated by x and y, and their relations are x^3 = e, y^2 = e, and yx = x^2y (since x^2 = z = y^(-1)xy). Now, let D3 be the dihedral group of order 6, generated by elements a and b with relations a^3 = e, b^2 = e, and ba = a^2b. Define a mapping φ: G → D3 such that φ(x) = a and φ(y) = b. We will show that φ is an isomorphism. First, we show that φ is a homomorphism. For any g, h in G, we have: 1. If g = x^i and h = x^j, where i, j ∈ {0, 1, 2}, then φ(gh) = φ(x^(i+j)) = a^(i+j) = a^i * a^j = φ(g)φ(h). 2. If g = x^i and h = yx^j, where i, j ∈ {0, 1, 2}, then φ(gh) = φ(yx^(i+j)) = b * a^(i+j) = a^i * b * a^j = φ(g)φ(h). 3. If g = yx^i and h = x^j, where i, j ∈ {0, 1, 2}, then φ(gh) = φ(yx^(i+j)) = b * a^(i+j) = a^(-i) * b * a^i * a^j = φ(g)φ(h). 4. If g = yx^i and h = yx^j, where i, j ∈ {0, 1, 2}, then φ(gh) = φ(x^(i+j)) = a^(i+j) = a^i * a^j = φ(g)φ(h). Thus, φ is a homomorphism. Next, we show that φ is bijective. Since |G| = |D3| = 6, it is enough to show that φ is either injective or surjective. To show that φ is injective, suppose φ(g) = φ(h) for some g, h in G. Then, g and h must have the same form (either both are powers of x or both are products of y and a power of x). Moreover, since φ is a homomorphism, g and h must have the same exponent for x. Thus, g = h, and φ is injective. Since φ is a homomorphism and injective, it is also surjective, as |G| = |D3|. Therefore, φ is a bijection. Since φ is a bijective homomorphism, it is an isomorphism. Thus, G is isomorphic to D3.

answer:To determine whether G is isomorphic to C6 or D3, we will use the following facts: 1. C6 is a cyclic group of order 6, generated by a single element of order 6. 2. D3 is a dihedral group of order 6, which can be thought of as the symmetries of an equilateral triangle. It has one element of order 3 (a rotation) and three elements of order 2 (reflections). Now, let's analyze the group G. Since G has order 6, by Lagrange's theorem, the possible orders of elements in G are 1, 2, 3, and 6. First, let's check if G has an element of order 6. If it does, then G is generated by that element and is isomorphic to C6. If G does not have an element of order 6, then we will check the structure of G to see if it matches the structure of D3. Case 1: G has an element of order 6. Let g be an element in G with order 6. Then, G = {e, g, g^2, g^3, g^4, g^5}, where e is the identity element. Since G is generated by a single element of order 6, G is isomorphic to C6. Case 2: G does not have an element of order 6. In this case, we know that G must have elements of orders 1, 2, and 3. Let's analyze the structure of G: - There is exactly one element of order 1, which is the identity element e. - Let r be an element of order 3. Then, G contains {e, r, r^2}. Since r^3 = e, r^4 = r, and r^5 = r^2, there are no other elements of order 3. - Let s be an element of order 2. Then, G contains {e, s}. Since G has order 6, there must be two more elements of order 2, say s' and s''. Now, G = {e, r, r^2, s, s', s''}. Notice that this structure matches the structure of D3, with one element of order 3 (r) and three elements of order 2 (s, s', s''). Therefore, G is isomorphic to D3. In conclusion, to determine whether G is isomorphic to C6 or D3, we first check if G has an element of order 6. If it does, then G is isomorphic to C6. If not, then G must have the structure of D3, and G is isomorphic to D3.

answer:To determine the number of distinct isomorphism classes of groups of order 6, we will use the following facts: 1. A group of order 6 must have a subgroup of order 2 and a subgroup of order 3, by Lagrange's theorem. 2. If a group has a normal subgroup of order 3 and a normal subgroup of order 2, then the group is a direct product of these subgroups. Now, let G be a group of order 6. By fact 1, G has a subgroup H of order 3 and a subgroup K of order 2. Since the index of H in G is 2, H is a normal subgroup of G. Similarly, since the order of K is 2, K is also a normal subgroup of G. Now, let's consider the two possible cases: Case 1: G is a cyclic group of order 6. In this case, G is isomorphic to the cyclic group C6, which is generated by an element of order 6. This group has a unique subgroup of order 3 and a unique subgroup of order 2. Case 2: G is not a cyclic group. In this case, G cannot have an element of order 6. Since G has a normal subgroup H of order 3 and a normal subgroup K of order 2, by fact 2, G is isomorphic to the direct product H x K. The direct product of a group of order 3 and a group of order 2 is the dihedral group D3, which is the group of symmetries of an equilateral triangle. Therefore, there are only two distinct isomorphism classes of groups of order 6: the cyclic group C6 and the dihedral group D3. Every group of order 6 is isomorphic to either C6 or D3.

answer:To prove that G is isomorphic to either C6 or D3, we will use the following facts: 1. If a group has an element of order equal to its size, then it is a cyclic group. 2. If a group has a subgroup of index 2, then it has a normal subgroup of index 2. First, let's consider the possible orders of elements in G. By Lagrange's theorem, the order of an element must divide the order of the group. Since the order of G is 6, the possible orders of elements in G are 1, 2, 3, and 6. If G has an element of order 6, then G is a cyclic group generated by that element, and G is isomorphic to C6. Now, let's assume that G does not have an element of order 6. Then, the possible orders of elements in G are 1, 2, and 3. Since the identity element has order 1, there must be at least one element of order 2 and one element of order 3. Let a be an element of order 2 and b be an element of order 3. Then, a^2 = e and b^3 = e, where e is the identity element. Consider the subgroup H = {e, a}. Since a has order 2, H has order 2. Thus, H is a subgroup of index 2 in G. By fact 2, G has a normal subgroup of index 2, which must be H. Now, consider the cosets of H in G. Since H is a normal subgroup, the left and right cosets are equal. There are two cosets: H itself and bH = {b, ba}. Since G has 6 elements and H has 2 elements, the cosets partition G into two sets of 3 elements each. Now, let's consider the product of elements from different cosets. For any h1, h2 in H and c1, c2 in bH, we have: (h1c1)(h2c2) = h1(c1h2)c2 Since H is a normal subgroup, c1h2 is in H, so the product is in bH. This shows that the product of elements from different cosets is again in bH. Finally, let's consider the product of elements within the coset bH. For any c1, c2 in bH, we have: c1c2 = (bh1)(bh2) = b(h1h2)b^(-1) Since h1h2 is in H, the product is in H. This shows that the product of elements within the coset bH is in H. Thus, G satisfies the defining properties of the dihedral group D3: 1. G has a normal subgroup of index 2 (H). 2. The product of elements from different cosets is in bH. 3. The product of elements within the coset bH is in H. Therefore, G is isomorphic to the dihedral group D3. In conclusion, G is isomorphic to either the cyclic group C6 or the dihedral group D3.