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I've come across an alternate definition of a group object, shown below:

Suppose $F:C^{op} \rightarrow Grp$ is a functor such that its composition with the forgetful functor $?:Grp \rightarrow Set$ is representable by some object $X \in C$. X is called a group object in C.

You can see more on this definition here and here. Now, I see the intuition and equivalence, on how to derive the usual notion of a group object from the above. But this definition by itself is strange to me. How can a group object just be an object in the category, with no associated morphisms? More specifically, what I have been struggling with is the following: it is an elementary fact that a group object in $Sets$ is a group, and a group object in $Grps$ is an abelian group. I know how to show this using the conventional definition of a group object, but how does one show it for this definition, without resorting to the usual definition?

My vague idea: Call $G$ the group object and let $X \in C$. Every $Mor(X, G)$ can be interpreted as a group. In particular this will help us identify our identity (when $C = Set$) by looking at $Mor(\{*\},G)$, and my overall impression is that I should be using the morphisms into G and their associated group structure to uncover a group structure on $G$. However, I am still made uneasy by the fact that $G$ is just an object with no more structure. For instance, on $Set$, G is just a set, and picking out an element and saying "this will act as the identity" doesn't mean much when there is no multiplication or inverse or anything encoded into $G$.

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If you understand that the definitions are equivalent, then you understand that $G$ actually does come equipped with structure arising from the lifting of its eepresentable functor to groups.

For the specific question about group objects in groups, let $G$ be a group object with multiplication $*$ and let $\times$ be the multiplication on $Hom(X,G)$ determined by the group object structure on $G$. Then $*$ induces another multiplication on $Hom(X,G)$ which is a homomorphism with respect to $\times$, so the Eckmann-Hilton argument applies to show that $\times=*$ and that both operations are commutative.

EDIT: To expand on the above, by the Yoneda lemma, one gets a group homomorphism $\mu:G\times G\to G$ mapped to the multiplications on $Hom(Z,G)$ by the Yoneda embedding. Furthermore, one gets a unit $e:1\to G$ mapped to the units of $Hom(Z,G)$ by Yoneda. The Eckmann-Hilton argument shows that $\mu$ and the given multiplication of $G$ must coincide and commute.

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  • $\begingroup$ I understand everything except how the new multiplication is induced, and how it is a homomorphism. Any help? $\endgroup$ Nov 26 '19 at 16:13
  • $\begingroup$ Alright, here's where I'm at right now. I have a group $(G, *)$ and a group $(Hom(X, G), \circ)$. I want a new operation on $Hom(X,G)$ that will help me do Eckmann-Hilton, but I can't find it. I've been trying to pointwise multiplication using $*$ but it just doesn't work as a homomorphism with respect to $\circ$. I've been tearing my hair out here so I would really appreciate some help. $\endgroup$ Nov 26 '19 at 17:23
  • $\begingroup$ @Confused_Undergrad hope my expansion helps some. $\endgroup$ Nov 27 '19 at 0:02

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