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Is it possible to define a group as a category?

What exactly will be objects of this category and how will we say that every element should have an inverse?

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Do you mean that the group itself is the category? That is, the elements of the groups are the objects of the category? Or do you mean the category of groups? – Asaf Karagila Nov 4 '11 at 16:35
I mean that the group itself as a category not the category of groups. – Mohan Nov 4 '11 at 16:37
This answer contains, inter alia, a description of how to view a group as a category (also, how to view a monoid as a category, and how to view a set as a category). – Arturo Magidin Nov 4 '11 at 17:07
up vote 5 down vote accepted

Yes, you can. Define a category $M$ with just one formal object say $ob(M) = \{X\}$. Let $G$ be a group. Define $Mor(X,X)$ = underlying set of $G$, and composition of morphisms in $Mor(X,X)$ by the binary operation on $G$. The identity morphism on X is just the identity element in $G$. Then you can verify that all axioms of a category are satisfied by $M$. Since each element in $G$ has an inverse, note, moreover, that every element in $Mor(X,X)$ is an isomorphism. In fact, you can define any monoid as a category with just one formal object in this way. But, with a monoid viewed as a category with one object, it is no longer true that every morphism of that object is an isomorphism.

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There is an isomorphism (of categories) from the category of groups to the category of 1-groupoids.

A 1-groupoid is a category with precisely one object $X$ with the property that all morphisms $f$ in $\mathrm{Hom}(X,X)$ are isomorphisms.

It is not hard to write down the functor from the category of groups to the category of 1-groupoids mentioned above.

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Careful, I've never seen the term "1-groupoid" to refer to what you describe. In (possibly) more common terminology, a 1-groupoid is just a groupoid, i.e. the n=1 case of an n-groupoid. – Santiago Canez Nov 4 '11 at 17:58
Right, I should have said that. An n-groupoid is defined similarly. A category with n objects such that all morphisms are isom's. (or something like that.) – Hannah Nov 5 '11 at 9:37
@Hannah: that is not what the term $n$-groupoid usually means. See for example. – Qiaochu Yuan Mar 5 '12 at 5:27
moreover, there is no isomorphism between the two categories (that of groups and that of one-object categories with all arrows invertible), only an equivalence. The reason is that a category with one object still needs to name that one object. – Ittay Weiss Dec 30 '12 at 1:19

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