Reading about group objects in categories, it's a fact that a group object is in the category of $\mathcal{Set}$ just a common group. I am trying to give an actual proof of this, but I'm a bit confused on how to connect the group axioms to the definition of a group object - mainly because the group axioms work with elements inside of a group, but in category theory, the elements of an object don't really play a role. So I wonder how to actually proof that a group object in $\mathcal{Set}$ is indeed a group.

The definition of a group object as I know is the following, given by Richard Pink in "Finite group schemes":

A (commutative) group object in the category $\mathcal{C}$ is a pair consisting of an object $G \in ob(\mathcal{C})$ and a morphism $\mu : G \times G \to G$ such that for any object $Z \in ob(\mathcal{C})$ the map $G(Z) \times G(Z) \to G(Z)$, $(g, g') \mapsto \mu \circ (g, g')$ defines a (commutative) group.

Where $G(Z)$ is the set of morphisms $Z \to G$.

Im also not sure if this is a proper definition (e.g. shouldn't it be explained what is ment by "define a group").

  • $\begingroup$ Here "defines a group" means that the set $G(Z)$ is a group under the binary operation $g\times g'=\mu\circ(g,g')$. (The notation is a bit confusing here; the first $(g,g')$ is a pair of morphisms that you're multiplying to define the group structure on $G(Z)$, whereas the $(g,g')$ in $\mu\circ(g,g')$ is the pair considered as a single map $Z\to G\times G$, by $z\mapsto(g(z),g'(z))$). $\endgroup$ – mdp Oct 24 '14 at 14:47
  • $\begingroup$ Thank you for the clarification. But what about the first confusion? My assumption is, that for showing, that something is a group, I have to show the group axioms. But they are defined on elements of a group. But how to relate elements to objects in categories - since nowhere in category theory so far I've seen that the elements of an object matter? $\endgroup$ – cbb Oct 24 '14 at 15:07
  • $\begingroup$ As pointed out in the answer, the points of a set $G$ can be naturally identified with the maps $X\to G$ where $X$ is any one-point set. $\endgroup$ – mdp Oct 24 '14 at 15:18
  • $\begingroup$ It is also possible to phrase the set-theoretic definition of a group in terms of maps $\mu\colon G\times G\to G$ (the multiplication), $\varepsilon\colon\{e\}\to G$ (the unit map, whose image is the identity) and $i\colon G\to G$ (inversion) which are required to satisfy various identities (usually expressed as commutative diagrams). This gives an alternative (and presumably equivalent!) definition of a group object in a category. $\endgroup$ – mdp Oct 24 '14 at 15:21

Take a group $G$. Then, pointwise multiplication certainly embeds $\hom_{\mathsf{Set}}(Z,G)$ with a structure of group for any set $Z$.

Conversely, if $G$ is group object of set, you can take advantage of the fact that $$ G \simeq \hom_{\mathsf{Set}}(1,G) $$ (where $1$ is a final object of the category $\mathsf{Set}$, that is a singleton).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.