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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  • $\begingroup$ 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. $\endgroup$ – user113102 May 16 at 16:20
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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 $$ m(i(a), a) = m \circ (i \times Id_G)(a, a) = m \circ (i \times Id_G) \circ \Delta(a) = e. $$
  • $\begingroup$ Thank you very much! $\endgroup$ – colbyjack101 May 16 at 19:39

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