# Group objects in category of groups

I am trying to show that a group object in the category of Groups is an abelian group (which, to a beginner, reads as a peculiar statement!)

So here is what I know:

Let $\mathscr{C}$ be a category having (finite) products and a terminal object $Z$. A group object in $\mathscr{C}$ is an object $G$ and morphisms $\mu: G \times G \to G$, $\eta:G \to G$ and $\epsilon: Z \to G$ such that the diagrams for associativity, identity and inverse commute (apologies, it is too hard to draw them without xymatrix here)

Here I think of (hopefully correctly), $\eta$ as the inversion $g \mapsto g^{-1}$

In the category of groups we have that the terminal object is any trivial group, which I will just call $0$.

Then to show the result, I would need to take $g_1,g_2 \in G$ and show that $\mu(g_1,g_2) = \mu(g_2,g_1)$

I am a bit unsure where to go from here. The problem, I guess, is that I don't see what makes the category of groups lead to abelian group objects. For example, in the category of Sets the group objects are just groups.

The main point is that you require the inverse $\eta$ to be a group homomorphism (i.e. a morphism in the category of groups). You can easily check that this forces $G$ to be abelian, using the compatibility between multiplication $\mu$ and inversion $\eta$ (I will use the usual group notation, you can convert it into $\mu$-$\eta$-ology): $(gh)^{-1} = h^{-1}g^{-1}$, and that is supposed to be equal to $g^{-1}h^{-1}$ by the requirement that $\eta$ is a morphism.
Another issue is: why does the morphism $\mu$ have to be the group structure on $G$ that already comes from $G$ being an element of $\textbf{Grp}$? This is known as the Eckmann-Hilton argument.
• @Alex: so in the notation of the question it boils down to: $\mu \eta(g,h) = \mu(\eta(h),\eta(g))=\mu(\eta(g),\eta(h))$? Commented Apr 18, 2011 at 3:25
• @Omar I think you do need Eckmann-Hilton to finish the argument, as you say. That's the reason why I included it. But you can prove that the structure "$*$", to use your terminology, gives an abelian group, as in the first paragraph. Then, as you say, you need Eckmann-Hilton to show that "*" and "." coincide. Commented Apr 19, 2011 at 6:08