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 $K$, $\alpha \circ \theta _{1}=\alpha \ circ\theta _{2}\implies \theta _{1}=\theta _{2}$.

Related facts

• Epimorphism iff surjective in the category of groups

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.