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$\begingroup$

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.

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    [Eckmann-Hilton](http://en.wikipedia.org/wiki/Eckmann%2DHilton%5Fargument)2011-04-18

1 Answers 1

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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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    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.2011-04-18
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    @Mariano: yes I believe that's what I do in the first paragraph. Or did you mean something different?2011-04-18
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    @Alex, Indeed, that is why I wrote "indeed" :) I was just trying to emphasize that E-H is not needed.2011-04-18
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    @Mariano Got you. In that case a +1 for your comment :-)2011-04-18
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    @Alex: so in the notation of the question it boils down to: $\mu \eta(g,h) = \mu(\eta(h),\eta(g))=\mu(\eta(g),\eta(h))$?2011-04-18
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    @Qwirk The first term should read $\eta\mu(g,h)$ (otherwise you are feeding two arguments into a unary operator) but other than that, that's indeed the translation of what I wrote into $\mu$-$\eta$-ology.2011-04-18
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    @Alex yes, that is what I meant - for some reason I can't edit comments for very long!2011-04-18
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    @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.2011-04-19
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    @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.2011-04-19
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    @Alex: I still don't understand. I think when you write $(gh)^{-1} = h^{-1} g^{-1}$ you probably mean the multiplication and inverse map giving $G$ the structure of a __group object__ in Grp, but that when you write $(gh)^{-1} = g^{-1} h^{-1}$ you mean the same inverse as before but now need the multiplication $G$ has a an __object__ of Grp. I think denoting both multiplications as $gh$ tripped you up. Am I missing something?2011-04-19
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    @Omar You are right, you first need to apply the first half of Eckmann-Hilton (the bit about the two structures coinciding), before you can use the argument in the first paragraph.2011-04-20