If ℤⁿ≅ℤᵐ then n=m.

Let ψ:ℤⁿ—>ℤᵐ a group isomorphism. Let’s prove that n=m.Let:

H=2ℤⁿ={(2z1,…,2zn) : zi∈ℤ ∀ i=1,..,n} ≤ ℤⁿ
K= 2ℤᵐ={(2z1,…,2zm) : zi∈ℤ ∀ i=1,..,m} ≤ ℤᵐ

Observe that ψ(H)=K. Indeed if (2z1,…,2zn) ∈ H then ψ (2z1,…,2zn) = 2ψ(z1,…,zn). Viceversa if (2z1,…,2zm) ∈ K then ψ⁻¹(2z1,…,2zm)∈H —> (2z1,…,2zm) ∈ψ(H).

Now, since ℤᵐ e ℤⁿ are abelian H and K are normal subgroups then we can consider ℤⁿ/H, ℤᵐ/K. Let’s define:

φ: ℤⁿ/H —> ℤᵐ/K, (z1,…,zn)+H |—> ψ(z1,…,zn)+K

• φ is well defined . Indeed if (z1,…,zn)+H = (z’1,…,z’n)+H then (z1,…,zn)-(z’1,…,z’n) = (2k1,..,2kn) which imply φ( (z’1,…,z’n) + H) = φ ( (z1,…,zn) – (2k1,…,2kn) + H) = φ( (z1,…,zn) + H ) .
• φ is a homomorphism . This is trivial since ψ is a homomorphism.
• φ is bijective .In fact it is surjective because ψ is and it is injective because if ψ(z1,…,zn)+K = K then (z1,…,z )∈ H.

It follows that φ is an isomorphism . Now:

ℤⁿ/H ≅ (ℤ₂)ⁿ e ℤᵐ/K ≅ (ℤ₂)ᵐ

by the maps

(z1,…,zn)+H |—> ([z1]₂,…,[zn]₂)
(z1,…,zm)+K |—> ([z1]₂,…,[zm]₂)

then:

2ⁿ=| (ℤ₂)ⁿ | = | (ℤ₂)ᵐ | = 2ᵐ —> n=m.