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Suppose we have the category with only one object - the group G.

Why can we think of the morphisms in this category as of the elements of the group G?

I would be very grateful for explanation.

user1729
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DeltaDelta
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2 Answers2

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In general the morphisms of a category with one object form a monoid. The existence of inverses isn't guaradnteed. But any group can be viewed as a category with one object (again, actually any monoid can). Let us call the object $A$, it doesn't really matter. By definition of a category the morphism set $\text{Hom}(A,A)$ has a binary operation given by composition. Again by assumptions of a category, this composition is associative, and there is a two-sided identity. So it is a monoid. In particular if we have some monoid $M$ in mind we may take the category with $\text{Hom}(A,A) = M$ and where the composition of two morphisms is simply their product in the monoid.

A group has inverses, so a group will give you a category in which all the morphisms are invertible.

jxnh
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    But as always in category theory, arrows are important aswell and what makes the abovementioned association between categories and monoids fruitful is the fact that monoid homomorphisms correspond one-to-one to functors between one-object-categories. Thus, we can think of $\mathbf {Mon} $ as a full subcategory of $\mathbf{Cat} $. –  Aug 23 '14 at 18:35
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As pointed out by JHance, the endomorphisms of the unique object of a one-object-category do in general only form a monoid.

If you start with a group $G$, you can consider the category with one element that has as endomorphisms the elements of $G$ and composition of morphisms in this category is given by multiplication in the group $G$.

A natural generalization of this situation to categories with more than one element are groupoids. A groupoid is a category in which every morphism is an isomorphism. If there is only one object in the category, this is nothing but a group.

user44400
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