# Alternate definition of Group Object, and how to 'play' with it?

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 $$G$$.

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.
• 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. Nov 26 '19 at 17:23