AB-V2-30

Let me work through this formally. **Symbolic setup:** - G₁ = (ℤ, +) — the integers under addition - G₂ = (ℤ/7ℤ, +₇) — integers mod 7 under addition **Key invariant: cardinality is preserved under isomorphism.** An isomorphism φ: G₁ → G₂ must be a **bijection** (one-to-one and onto) that preserves the group operation. A bijection requires |G₁| = |G₂|. - |G₁| = |ℤ| = ℵ₀ (countably infinite) - |G₂| = |ℤ/7ℤ| = 7 Since ℵ₀ ≠ 7, no bijection φ: ℤ → ℤ/7ℤ can exist. Without a bijection, there is no isomorphism. Period. **Why "both cyclic" is insufficient:** Being cyclic is a necessary condition for two cyclic groups to be isomorphic, but not sufficient. The classification theorem for cyclic groups states: - Every infinite cyclic group is isomorphic to ℤ - Every finite cyclic group of order n is isomorphic to ℤ/nℤ - ℤ ≇ ℤ/nℤ for any n Cardinality is absolutely a structural (algebraic) property — it is invariant under isomorphism. **The answer is (B):** No — an infinite group cannot be isomorphic to a finite group because no bijection exists between them.

**Applying reasoning injection before solving:** **Answer: (B)** Here's the step-by-step reasoning path, checking each intermediate state: **Step 1 — Start state:** Two cyclic groups, both generated by 1. **Step 2 — Definition of isomorphism (the grounding axiom):** Two groups G and H are isomorphic iff there exists a **bijection** φ: G → H that preserves the group operation: φ(a + b) = φ(a) + φ(b). This definition is universal and unambiguous — option (D) is false. **Step 3 — Feasibility check on bijection:** A bijection requires equal cardinality. - |ℤ| = ℵ₀ (countably infinite) - |ℤ/7ℤ| = 7 (finite) These cardinalities differ. No bijection can exist between them. This is not a definitional question — it's a set-theoretic fact that terminates the argument cleanly. **Step 4 — Verify via homomorphism trace (falsification):** If φ: ℤ → ℤ/7ℤ were a homomorphism, then φ(7) = 7·φ(1) ≡ 0 (mod 7). But 7 ≠ 0 in ℤ, so φ(7) = φ(0) = 0 — meaning φ is not injective. No bijective homomorphism exists. **Step 5 — Close the path:** Options (A) and (C) commit the "start-end-only" error — they observe that both groups are cyclic and jump to "same structure," skipping the intermediate feasibility constraint (equal cardinality). "Cyclic" describes a *type* of group, not a unique isomorphism class. The isomorphism classes are: one for ℤ, and one for each ℤ/nℤ. They are all distinct. **The correct answer is (B).** An infinite group and a finite group cannot be isomorphic because isomorphism requires a bijection, and no bijection exists between sets of different cardinalities. Size *is* a structural invariant — isomorphic groups must have the same order.