Let G and H be two groups. The groups G x Hand H x G are isomorphica)f...
Explanation:
To prove the correct answer, we need to show that if G × H and H × G are isomorphic, then G and H are isomorphic.
Proof:
Let f: G × H → H × G be an isomorphism between the two groups.
Step 1: G is isomorphic to HConsider the map π₁: G × H → G defined by π₁(g, h) = g. This map is a projection onto the first coordinate.
Similarly, consider the map π₂: H × G → G defined by π₂(h, g) = g. This map is a projection onto the second coordinate.
Since f is an isomorphism, it preserves the group structure. Therefore, the composition of f with the projections π₁ and π₂ gives:
f ◦ π₁: G × H → H × G → G
f ◦ π₂: H × G → G × H → H
Step 2: f ◦ π₁ and f ◦ π₂ are isomorphismsWe need to show that f ◦ π₁ and f ◦ π₂ are isomorphisms.
Step 3: f ◦ π₁ is an isomorphismTo prove that f ◦ π₁ is an isomorphism, we need to show that it is a bijection and preserves the group operation.
Injectivity: Suppose (g₁, h₁) and (g₂, h₂) are two elements in G × H such that f ◦ π₁(g₁, h₁) = f ◦ π₁(g₂, h₂). This implies that f(g₁) = f(g₂). Since f is an isomorphism, it is injective, so g₁ = g₂. Thus, (g₁, h₁) = (g₂, h₂), and f ◦ π₁ is injective.
Surjectivity: Let (h, g) be an element in H × G. Consider the element (g, h) in G × H. Since f is surjective, there exists an element (g, h) in G × H such that f(g, h) = (h, g). Therefore, f ◦ π₁ is surjective.
Preservation of the group operation: Let (g₁, h₁) and (g₂, h₂) be two elements in G × H. Then:
f ◦ π₁((g₁, h₁) · (g₂, h₂)) = f ◦ π₁(g₁g₂, h₁h₂) = f(g₁g₂) = f(g₁) · f(g₂) = f ◦ π₁(g₁, h₁) · f ◦ π₁(g₂, h₂)
Therefore, f ◦ π₁ preserves the group operation.
Step 4: f ◦ π₂ is an isomorphismUsing similar arguments as in Step 3, we can show that f ◦ π₂ is an isomorphism.
Step 5: G is isomorphic to HSince f ◦ π₁ and f ◦ π₂ are isomorphisms,