# Proving group objects in the category of sets are group [closed]

I'm having difficulty proving that in the category of sets, the group objects are just groups. I know that for a category, C with finite products, an object,G, is a group object with the morphisms m,e,i where m is associative and e is the identity morphism and i is the inversion morphism. Any help with this would be greatly appreciated.

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• Should just be that given a group object in Set, if you look at what $m$ does element wise, all of the facts about $m,e,i$ should give that $m$ defines an operation on the set that makes it into a group. – user113102 May 16 at 16:20
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## 1 Answer

To check that it is a group, you can simply check all the group axioms. The map $$e: 1 \to G$$ is just some element in $$G$$ since $$1$$ is a singleton in $$\mathbf{Set}$$. So we might as well denote this element by $$e$$ too. Then checking the axioms is pretty much direct from the commutativity of the relevant diagrams:

• Associativity. That $$m$$ is associative means that $$m \circ (m \times Id_G) = m \circ (Id_G \times m)$$, so we have for all $$a, b, c \in G$$, or equivalently $$(a,b,c) \in G \times G \times G$$, that $$m(m(a,b), c) = m \circ (m \times Id_G)(a,b,c) = m \circ (Id_G \times m)(a,b,c) = m(a,m(b,c)),$$ and so the multiplication is associative.
• Identity. The category-theoretic version says $$m \circ \langle Id_G, e \rangle = Id_G = m \circ \langle e, Id_G \rangle$$. So for any $$a \in G$$ this indeed gives us $$m(a, e) = m \circ \langle Id_G, e \rangle(a) = a = m \circ \langle e, Id_G \rangle(a) = m(e, a).$$
• Inverse. Note that since we always have an arrow $$G \to 1$$, we can compose this with $$e: 1 \to G$$ to get an arrow $$G \to G$$ that we will also denote by $$e$$ (bit of abuse of notation, but it is simply the map that sends everything to the identity element). On the category-theoretic side we have this time $$m \circ (Id_G \times i) \circ \Delta = e = m \circ (i \times Id_G) \circ \Delta$$ where $$\Delta: G \to G \times G$$ is the diagonal arrow. Which for any $$a \in G$$ translates to $$m(a, i(a)) = m \circ (Id_G \times i)(a, a) = m \circ (Id_G \times i) \circ \Delta(a) = e,$$ and $$m(i(a), a) = m \circ (i \times Id_G)(a, a) = m \circ (i \times Id_G) \circ \Delta(a) = e.$$
• Thank you very much! – colbyjack101 May 16 at 19:39