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.

  • $\begingroup$ 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

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.