Monomorphism iff injective in the category of groups

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This article gives a proof/explanation of the equivalence of multiple definitions for the term injective homomorphism
View a complete list of pages giving proofs of equivalence of definitions

Statement

The following are equivalent for a homomorphism of groups \alpha: H \to G:

  1. \alpha is injective as a set map.
  2. \alpha is a monomorphism with respect to the category of groups: For any homomorphisms \theta_1,\theta_2:K \to H from any group K, \alpha \circ \theta_1 = \alpha\ circ \theta_2 \implies \theta_1 = \theta_2.

Related facts

Proof

Injective homomorphism implies monomorphism

This follows simply by thinking of the maps as set maps. In general, for any concrete category, any injective homomorphism is a monomorphism.

Monomorphism implies injective homomorphism

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Given: \alpha is a monomorphism: For any homomorphisms \theta_1,\theta_2:K \to H from any group K, \alpha \circ \theta_1 = \alpha \circ \theta_2 \implies \theta_1 = \theta_2.

To prove: \alpha is injective, i.e., the kernel of \alpha is the trivial subgroup of G.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Let K be the kernel of \alpha. Let \theta_1 be natural inclusion of K in H and \theta_2 be the trivial homomorphism from K to H.
2 \alpha \circ \theta_1 = \alpha \circ \theta_2 as homomorphisms from K to G, since both \alpha \circ \theta_1 and \alpha \circ \theta_2 are trivial homomorphisms from K to G. Step (1) By Step (1), K is the kernel of \alpha and \theta_1 is its natural inclusion in H, so the composite is trivial. \alpha \circ \theta_2 is trivial because \theta_2 is trivial.
3 \theta_1 = \theta_2. \alpha is a monomorphism Step (2) Step-given combination direct.
4 K is trivial, i.e., the kernel of \alpha is trivial. Steps (1), (3) Step (3) tells us that the inclusion of K in H is the same as the trivial homomorphism, forcing K to be trivial. This completes the proof.