monomorphisms in the category of groups

This page is to record a constructive proof of the following result.

Every monomorphism in the category of groups is an equalizer.

Let $i: H \to G$ be monic, and let $\pi: G \to G/H$ be the canonical surjective function $g \mapsto g H$ . Let $A$ be the free abelian group on $G/H$ with $j: G/H \to A$ the canonical injection, and let $A^G$ denote the set of functions $f: G \to A$, with the pointwise abelian group structure inherited from $A$. This carries a $G$-module structure defined by

$(g \cdot f)(g') = f(g' g).$

For any $f \in A^G$, the function $d_f: G \to A^G$ defined by $d_f(g) = g f - f$ defines a derivation. Passing to the wreath product $A^G \rtimes G$, we have two homomorphisms $\phi, \psi: G \rightrightarrows A^G \rtimes G$ defined by $\phi(g) := (d_{j \pi}(g), g)$ and $\psi(g) := (0, g)$. I claim that $i: H \to G$ is the equalizer of the pair $\phi, \psi$. For,

$\array{
d_{j\pi}(g) = 0 & \text{iff} & (\forall_{g': G})\; g\cdot j\pi(g') = j\pi(g') \\
& \text{iff} & (\forall_{g': G})\; j\pi(g' g) = j\pi(g') \\
& \text{iff} & (\forall_{g': G})\; j(g' g H) = j(g' H) \\
& \text{iff} & (\forall_{g': G})\; g' g H = g' H \\
& \text{iff} & g H = H \\
& \text{iff} & g \in H.
}$

(All we needed was *some* injection $j: G/H \to A$ into an abelian group; I chose the canonical one.)

This proof can be adapted to show that monomorphisms in the category of finite groups (group objects in $FinSet$) are also equalizers. All that needs to be modified is the choice of $A$, which we could take to be the $\mathbb{F}_2$-vector space freely generated by $G/H$.

The category of groups is balanced: every epic mono is an isomorphism.

This follows because an epic equalizer is an equalizer of two maps that are the same, hence an isomorphism.

Every epimorphism in the category of groups is a coequalizer.

Since every morphism $f: G \to H$ factors as a *regular* epi $p: G \to G/\ker(f)$ followed by a mono $i$, having $f$ epic implies $i$ is a epic mono. Epic monos $i$ being isomorphisms, $f$ is then forced to be regular epic as well.

Revised on January 1, 2020 at 12:33:27
by
Todd Trimble