# Group objects in $\mathbf{Ab}$ and $\mathbf{Grp}$

Aluffi suggests (II.10.4) proving that every abelian group has exactly one structure of group object in the category $$\mathbf{Ab}$$, and asks about what would be a group object in $$\mathbf{Grp}$$ (II.10.5).

So, let's assume that $$\circ : G \times G \rightarrow G, \iota : G \rightarrow G, e : 1 \rightarrow G$$ are the corresponding elements of some group object over $$G \in \mathbf{Ab}$$ (if they exist, but the existence is easy), and all the diagram commutativity requirements hold for them. Also let's explicitly mention they are morphisms in $$\mathbf{Ab}$$ which implies they are abelian group homomorphisms.

Previous exercise that I worked out suggests proving that if $$(G, \cdot)$$ is a group and $$\circ : G \times G \rightarrow G$$ is a group homomorphism such that $$(G, \circ)$$ is also a group, then $$\circ$$ and $$\cdot$$ coincide. It feels natural to piggyback on it and just prove that $$(G, \circ)$$ has a group structure with the inverse being $$\iota$$ and identity being the element chosen by $$e$$ (which we'll also call $$e$$), and then the uniqueness immediately follows (and, moreover, it follows that the group object over $$G$$ is $$G$$ itself, in a sense).

So let's just check the group structure:

• $$g_1 \circ g_2$$ as a group operation on $$G$$ is well-defined and "typechecks", since $$G \times G$$ is just the cartesian product (with the right group structure on top) in $$\mathbf{Ab}$$.

• $$\forall g : G. e \circ g = g = g \circ e$$ — holds, since the left-identity diagram requires that $$\circ (e, g) = g$$, and similarly for the right-identity.

• $$\forall g : G. \exists g' : G. g'g = gg' = e$$ — satisfied by taking $$g' = \iota(g)$$ and using the commutativity of the diagrams involving $$\iota$$.

• Associativity holds similarly.

All good. $$(G, \circ)$$ has a group structure, so it coincides with $$(G, \cdot)$$, so we're done.

Now let's get to $$\mathbf{Grp}$$. Basically the same argument as above should also hold, since I don't think I've used any abelianness, so the group object in $$\mathbf{Grp}$$ over $$G$$ coincides with $$G$$ itself. But I know that group objects in $$\mathbf{Grp}$$ are abelian groups too, so where am I mistaken?

I noticed this question has been asked a few times here, but I'm rather curious about any holes in my argument than just getting the proof done.

• Google Eckmann-Hilton argument. Jun 27, 2019 at 20:32
• The exercise that I (try to) reduce the problem to is an an instance of that argument, isn't it? Jun 27, 2019 at 20:34
• Yes it is, and whatever proof you will come up with, it will be a variation on that argument. Jun 27, 2019 at 20:39

The inverse map $$\iota$$ is now not just a function, but a morphism in the category $$\mathbf{Grp}$$.
We already know that if $$g\cdot h$$ is an object in $$G$$, then $$(g\cdot h)^{-1} = h^{-1}\cdot g^{-1}$$. Because of your previous exercise (Eckmann-Hilton), this carries over to the group object: $$\iota(g\circ h) = \iota(h)\circ \iota(g)$$.
Now because $$\iota$$ is a group morphism, we also have $$\iota(g\cdot h) = \iota(g)\cdot \iota(h)$$. This additional property establishes that $$\iota(g)\circ \iota(h) = \iota(gh) = \iota(h)\circ \iota(g)$$, which proves commutativity.
Your proof was entirely correct; you just didn't use the fact that $$\iota$$ is a group morphism, which means that (because the multiplication structure is the same) the group object, and hence the original group, must be commutative. (Noncommutative groups don't have corresponding group objects because their inversion operator fails to be a homomorphism of groups.)
• That's the last part that I have trouble understanding. Let's say I have $g_1, g_2 \in G$. What happens to $g_1 \circ g_2$ if $g_1 g_2 \neq g_2 g_1$? Or does this mean that group objects only exist for those groups that are commutative? Jun 27, 2019 at 21:19
• I believe group objects only exist for commutative groups: Group objects in $\mathbf{Grp}$ are commutative (a group is commutative just if inversion is a group homomorphism), and by Eckmann-Hilton, $\cdot$ and $\circ$ coincide, so the original group $G$ itself must be commutative. Jun 27, 2019 at 22:18
• @0xd34df00d: Perhaps the observation $(ab)^{-1}=b^{-1}a^{-1}$ will clarify why inversion is not a morphism. Jun 28, 2019 at 1:44