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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]($h$ttp://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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    @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