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From MacLane's Category Theory:

A group is a category with one object in which every arrow has a two-sided inverse under composition.

So the analogies I see between a category group $CG$ and a group $G$ is as follows:

Arrows : Elements of the group

Identity arrow : identity element

composition of arrows : group operation

Associativity function : associative group operation

Unit law : identity properties

All composable arrows exist : closure of the group

But the one I can't seem to understand is: What is analog for the single object in $CG$?

Single object in category : (Something for a group)?

Community
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Oliver G
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2 Answers2

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While Arthur’s answer is correct, in most constructions the object isn’t anything. It’s just a formality introduced to give us somewhere to stick the morphisms. It’s a standard point in category theory that the objects aren’t where the content is, and that’s particularly true here.

Kevin Carlson
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  • This is the most useful answer. It's like asking what the single element of a terminal object $1$ in $\mathsf{Set}$ is: it could be anything, and it doesn't matter for what we want to do with it. – Mees de Vries Dec 17 '19 at 15:27
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The single object in this category could be the set of elements in the group, where each morphism corresponds to the function from the group to itself given by multiplying by an element of the group. That way both $\hom(G, G)$ and $G$ itself correspond to the elements of the group, but in different ways. That's the most natural thing that I can think of.

But at the end of the day, it's not really important exactly what this object is.

Arthur
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  • I'm a little confused by your description. You say "each morphism corresponds to the function from the group to itself by multiplying by an element of the group", but isn't each morphism (arrow) just an element of the group? If you're saying $a \in G$ is a function from $G$ to $G$ by some multiplication, what two things are you multiplying? – Oliver G Dec 16 '19 at 17:06
  • @OliverG I'm saying that the arrow $a:G\to G$, for any element $a$ in the group, is the function $g\mapsto ag$. – Arthur Dec 16 '19 at 17:14
  • What is $g$ in the function you just defined and what is its relation to the arrow a? – Oliver G Dec 16 '19 at 17:16
  • @OliverG $g$ is an arbitrary element of $G$. Given any group $G$, and any element $a\in G$, "multiplying by $a$" is a function from $G$ to $G$. It takes any $g\in G$ to the element $ag\in G$. In this case, $a$ (and the function mentioned above) corresponds to an arrow in the category, while the $g$'s are elements of the object. They are both elements of the group, but manifest differently in the category. – Arthur Dec 16 '19 at 17:19
  • So you're saying the analogy is (obj: set $G$, arrows : the function $a : G \rightarrow G$ by $g \mapsto ag$, identity arrow : $id_G : G \rightarrow G$ by $g \mapsto eg$ for group identity $e$, composition defined as $ba$ is the function $g \mapsto bag$) and etc for associativity and unit. Is that what you're saying? – Oliver G Dec 16 '19 at 17:30
  • @OliverG Yes, that is what I'm saying. – Arthur Dec 16 '19 at 19:26