nLab epimorphisms of groups are surjective

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Context

Group Theory

group theory

Classical groups

Finite groups

Group schemes

Topological groups

Lie groups

Super-Lie groups

Higher groups

Cohomology and Extensions

Related concepts

Contents

Statement

Proposition

Every epimorphism in the category of groups is surjective (a regular epimorphism). Equivalently, the category of groups is balanced (every monic epimorphism is an isomorphism).

There are probably many proofs of this statement. Some (for example, I think the one in Mac Lane's Categories for the Working Mathematician) involve some mild case analysis. The following (originally written up here) is more uniform (and constructive).

Proof

Suppose $i: H \hookrightarrow G$ is a monic epi. Let $A$ be a nontrivial abelian group, say $\mathbb{Z}/(2)$. There is a left action of $G$ on $A^G = \prod_{g \in G} A$, the group of functions $G \to A$, where $(g \cdot \phi)(g') \coloneqq \phi(g'g^{-1})$; this gives rise to a left $G$-module. It suffices to prove that any element $m \in A^G$ of the form

$G \stackrel{proj}{\to} G/H \to A$

is a constant function, i.e., is fixed under the action of $G$, or in other words that the map $\phi_m: G \to A^G$ mapping $g \in G$ to $g \cdot m - m$ is identically $0$. As $\phi_m$ is an $A^G$-valued derivation, it yields a splitting of the exact sequence of groups

$0 \to A^G \to G \ltimes A^G \to G \to 1$

given by $k: G \to G \ltimes A^G$, with $k(g) \coloneqq (g, \phi_m(g))$. Of course we also have the trivial splitting $j(g) \coloneqq (g, 0)$. But note that the restrictions along $i: H \hookrightarrow G$ coincide: $j \circ i = k \circ i$ since $\phi_m(h) = 0$ for all $h \in H$ (as $H$ acts trivially on the right on cosets $g H$). Since $i$ is epic, we conclude $j = k$, or that $\phi_m \equiv 0$, as was to be shown.

Last revised on November 6, 2020 at 18:09:20. See the history of this page for a list of all contributions to it.