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I've read two articles(1,2), but still have a question about structure of group, say $G$, as an one-object category. Let's call this category $C$.

  1. I understand that morphisms of $G$, which is denoted by $Mor(G,G)$, forms a monoid, which is explained in the first article(1). However, how can we say that $Mor(G,G)$ is a group? We just said $C$ has an object which is a group, not its morphism also should satisfy group's definition. How can we induce group properties directly from $C$'s definition?

  2. Also, I think it is natural to regard $Aut(G) \subseteq Mor(G,G)$, since every automorphism satisfy definition of a morphism. However, if we admit $Mor(G,G) \cong G$, then $Aut(G) \subseteq G$ can be nonsense, since we know the case that $G \cong Inn(G) \subset Aut(G) $, which implies $|G| < |Aut(G)|$.

What is wrong with my idea?

Community
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user124697
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    If a one-element category is a group, then every morphism is invertible. The more relaxed notion is that of a groupoid — a category all of whose morphisms are invertible, with no restrictions on the number of objects. – BrianO Feb 06 '16 at 02:46
  • @BrianO My question is that why the reason every morphism should be invertible. We just add two conditions in the category $C$ that it has only one object (which enforces $Mor(G,G)$ should satisfy definition of monoid), and that the object is group. I still have confused that why the condition (the object is group) enforces every morphism is invertible. – user124697 Feb 06 '16 at 02:49
  • Samuel Yusim's answer should address any remaining confusion (yes?) – BrianO Feb 06 '16 at 02:59
  • @BrianO Yes. thank you for my understanding on the category :) – user124697 Feb 06 '16 at 03:00
  • Possible duplicate of Group as a category – Alex M. Nov 05 '16 at 15:18

2 Answers2

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A one-object category is not necessarily a group. A one-object category where every morphism is invertible corresponds to a group.

This seems to be your misconception as far as I can tell, and it resolves both of your problems. Regarding (1), $C$ doesn't necessarily have an object which is a group. When every morphism is invertible, the set of morphisms form the group. Heck, you can formally define the object to be anything you like, but keep the same morphisms by how they compose, and you'll get the same group. Remember that the objects in a category aren't necessarily sets.

Regarding (2), once we require that all morphisms be invertible, you just get that your set of automorphisms is the set of morphisms, so your inclusions all collapse as needed.

EDIT: just some intuition, but this should say to you that a category is like a group, but we generalize the notion of a group element with the arrow, and the group operation with the notion of composition.

syusim
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The answer to your first question is no: as you've observed, we only need that $\operatorname{Hom}_{C}(G,G)$ is a monoid. You could take it to be the free monoid on one generator, which is just $\mathbb{N}$. (This is sort of a general model of a discrete dynamical system.)

I think the confusion in your second question arises from a type error. You're letting $G$ name the single object in your category, but you're also letting $G$ name the abstract group which is that single object's automorphism group.

Every automorphism is a morphism. In fact, in any category $\mathbf{C}$, $\operatorname{Aut}_{\mathbf{C}}(X) \subseteq \operatorname{Hom}_{\mathbf{C}}(X)$ for any object $X$ of $\mathbf{C}$ as monoids.

(Also, note that if $G$ is abelian, it has only one inner automorphism.)

mad_algebraist
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