TITLE: Group objects in category of groups
QUESTION [13 upvotes]: 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.

REPLY [16 votes]: 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.