# Monomorphism iff injective in the category of groups

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 $K$, $\alpha \circ \theta_1 = \alpha\ circ \theta_2 \implies \theta_1 = \theta_2$.

## 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.