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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").

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  • $\begingroup$ 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
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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).

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