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A typical example in category theory is to consider a given group $G$ as a category with one object. Then, $\hom(G, G)$, the set of arrows from $G$ to $G$, is defined to be the elements of $G$.

My problem is that I can't understand the relation between the set of arrows from $G$ to $G$ and the set of element of $G$. Is it that to each arrow is associated an element of $G$? How? It is hard for me to grasp the idea. Any help would be appreciated.

There is a question with a similar title here; but I couldn't find my answer there.

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Behrooz
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  • The set of arrows is $G$ itself. Do you know Cayley theorem? – user40276 Feb 14 '15 at 07:17
  • @ user40276. Unfortunately not... – Behrooz Feb 14 '15 at 07:22
  • Anyway, you just have to consider the category with one point and one invertible arrow for each $g \in G$, then composition is the multiplication operation in $G$. – user40276 Feb 14 '15 at 07:24
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    You'll get really confused if you label the object $G$. Label it something else, maybe $\bullet$. Then the definition is that $\text{Hom}(\bullet, \bullet)$ is $G$ and the composition map on homomorphisms is the multiplication in $G$. – Qiaochu Yuan Feb 14 '15 at 07:51

1 Answers1

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Maybe the situation becomes clearer when you consider the following statement:

A group "is the same thing" as a category with one object in which every morphism is an isomorphism.

In fact, if $G$ is a group, the corresponding category, say $C_G$, has one object $\bullet$ and the morphisms from $\bullet$ to itself are given by $G$, where the composition of two morphisms $\bullet \xrightarrow{g} \bullet$ and $\bullet \xrightarrow{h} \bullet$ is $\bullet \xrightarrow{gh} \bullet$.

Conversely, if $C$ is a category with one object $\bullet$ in which every morphism $f$ (necessarily from $\bullet$ to $\bullet$) is an isomorphism, then the set of morphisms from $\bullet$ to itself forms a group $G_C := Mor(\bullet, \bullet)$. The product of two group elements is given by the composition of morphisms. The unit element of the group is given by the identity morphism on $\bullet$ and the inverse of an element of a group is given by the inverse of the morphism (since they are all isomorphisms, this is always well-defined).

It is also worth noticing that a group is a special case of a groupoid (a category in which every morphism is an isomorphism). Namely, a group is a groupoid with one object.

Similarly, a monoid can be defined as a category with one object. It is a group if and only if every morphism is an isomorphism.

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