Final answer:
Any group homomorphism from a group G with prime order to another group G' must either be trivial or injective due to the structure of G and properties of the kernel of a homomorphism.
Step-by-step explanation:
To show that any group homomorphism ϕ: G→G' with the property that the order of the group G, denoted |G|, is a prime must either be the trivial homomorphism or a one-to-one map, we rely on the properties of group homomorphisms and the structure of groups of prime order.
Recall that a group homomorphism is a function between two groups that respects the group operations. That is, for all a, b in G, ϕ(ab) = ϕ(a)ϕ(b). The kernel of a homomorphism, Φ, is the set of all elements in G which are mapped to the identity element in G'. In formal terms, ker(ϕ) = g in G .
By the First Isomorphism Theorem, the image of G under ϕ, which is ϕ(G), is isomorphic to the quotient group G/ker(ϕ). Since |G| is prime, the only subgroups of G are the trivial subgroup {e}, where e is the identity in G, and G itself. Consequentially, the kernel of ϕ must likewise be one of these two subgroups.
If ker(ϕ) = G, then every element of G maps to the identity in G', making ϕ the trivial homomorphism. If ker(ϕ) = {e}, then ϕ is injective (one-to-one) because if ϕ(a) = ϕ(b), then ϕ(a)ϕ(b)^{-1} = e', implying that a*b^{-1} is in the kernel. But since the kernel is just {e}, this means a*b^{-1} = e, and hence a = b.