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.


1 Answer 1


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.

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    $\begingroup$ Notice that indeed one can prove commutativity without using the Eckmann-Hilton in this situation. E-H becomes useful when showing that a monoid in the category of monoids is abelian. $\endgroup$ Commented Apr 18, 2011 at 3:07
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    $\begingroup$ @Mariano: yes I believe that's what I do in the first paragraph. Or did you mean something different? $\endgroup$
    – Alex B.
    Commented Apr 18, 2011 at 3:16
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    $\begingroup$ @Alex: so in the notation of the question it boils down to: $\mu \eta(g,h) = \mu(\eta(h),\eta(g))=\mu(\eta(g),\eta(h))$? $\endgroup$
    – Juan S
    Commented Apr 18, 2011 at 3:25
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    $\begingroup$ @Mariano and @Alex: I don't understand this business about not needing Eckmann-Hilton here. You have one multiplication, say ".", on G because it is a group, and another, say "*", because it is a group object in the category of groups. The inverse is supposed to be a group homomorphism for the operation ".", but not a priori for "*". So the equations read (g*h)^(-1) = h^(-1)*g^(-1) (inverse!) and (g.h)^(-1) = g^(-1) . h^(-1) (homomorphism!). I don't see how you can finish the argument without basically doing Eckmann-Hilton. $\endgroup$ Commented Apr 19, 2011 at 2:38
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    $\begingroup$ @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. $\endgroup$
    – Alex B.
    Commented Apr 19, 2011 at 6:08

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