Understanding equivalence of category of group objects in the category of sets with category of Groups

Question

Let $\mathbf{C}$ be the category of group objects in the category of sets. We will show that $\mathrm{C}$ is equivalent to the category of Groups, say, $\textbf{D}.$

(a) Given an object in $\mathbf{C}$, describe the corresponding object in Groups

(b) Given an object in Groups, describe the corresponding object in $\mathbf{C}$.

(c) Conclude that there is a bijection $F: \mathrm{C} \rightarrow$ Groups between objects in $\mathrm{C}$ and objects in Groups.

(d) Given two objects $G, H$ in Groups and an arrow $\phi: G \rightarrow H$ in Groups, interpret $\phi$ as an abstract group homomorphism.

(e) Given two objects $G, H$ in $\mathrm{C}$ and an arrow $\phi: G \rightarrow H$ in $\mathrm{C}$, show that $\phi$ is just an ordinary group homomorphism.

(f) Conclude that there is a bijection between $\operatorname{Hom}_{\mathrm{C}}(G, H)$ and $\operatorname{Hom}_{\mathrm{Groups}}(F(G), F(H))$, where $\operatorname{Hom}_{\mathbf{D}}(A, B)$ is the set of arrows $A \rightarrow B$ in the category $\mathrm{D}$. We also call this bijection F. For any $\phi: G \rightarrow H$ in $\mathrm{C}$, we have $F(\phi): F(G) \rightarrow F(H)$ in Groups.

(g) Show that $F$ respects composition. That is, if $G, H, K \in \mathbf{C}$ and $\phi: G \rightarrow H$ and $\psi: H \rightarrow K$ are arrows, show that $F(\psi \circ \phi)=F(\psi) \circ F(\phi)$.

(h) Conclude that $\mathrm{C}$ and Groups are equivalent categories.

My try: a) Let $G$ be a group object in the category of sets, with corresponding multiplication, identity and inverse morphism as $m,t,i$ respectively. We will take the set $G$ and give group structure to it by defining $m(g,h):=g.h, t(*)=e, I(g)=g^{-1}.$ Here I am considering the terminal object in $\mathbf{C}$ as singleton $\{*\}$ and $e:\{*\} > \to G$. Then from the commutative diagrams it follows that $(G,.)$ is a group and hence, an element in $\textbf C$.

b) Completely the opposite. Let us say that $(G,.)$ is a group with identity $1$. Then, we will define that $m(g,h):=g.h, t(*)=e, > I(g)=g^{-1}.$ Here I am considering the terminal object in $\mathbf{C}$ as singleton $\{*\}$ and $e:\{*\} \to G$. Then we can check that $m,t,i$ are morphisms, which satisfies the commutative diagrams.

c) From our construction we are getting one to one map from both sides of the categories. Hence it will be a bijection.

I am stuck from here on. What is abstract group homomorphism and what does he mean by interpret? Is abstract group homomorphism in the category of group objects in the category of set? Please help me for the rest of the questions.

I am solving "http://pi.math.cornell.edu/~dmehrle/notes/old/pcmi15/groupobjectsetude.pdf" to get an understanding of group objects and stuck at the last question. As I didn't get any other undergrad level article on this. Solving these is providing a good insight. So please help.

One more thing: Won't $\textbf{C}$ be equivalent with category of Abelian groups?I don't have idea of limits and co-limits. I don't know whether that will be used here or not!

Answer 1

An 'abstract group homomorphism' in this context must be simply an arrow in $C$, that is, by definition of $C$, in case of $(G,m,t,i)$ and $(G',m',t',i')$ it's an arrow $\varphi:G\to G'$ in $Set$ that makes the following diagrams commutative: $$\matrix{G\times G&\overset m\longrightarrow & G\\ {\scriptstyle \varphi\times\varphi} \downarrow \phantom{\scriptstyle\varphi\times\varphi} && \phantom{\scriptstyle\varphi}\downarrow {\scriptstyle\varphi}\\ G'\times G' & \underset{m'}\longrightarrow & G'} \\ \matrix{1& \overset t\longrightarrow& G\\ \Vert && \phantom{\scriptstyle\varphi}\downarrow {\scriptstyle\varphi} \\ 1 & \underset{t'}\longrightarrow & G'} \quad \matrix{G& \overset i\longrightarrow & G\\ {\scriptstyle\varphi}\downarrow \phantom{\scriptstyle\varphi} && \phantom{\scriptstyle\varphi}\downarrow {\scriptstyle\varphi}\\ G'& \underset{i'}\longrightarrow & G'} $$

Verify that these mean exactly that $\varphi$ is a group homomorphisms with the corresponding group structures put on $G$ and $G'$.

It's not equivalent to the category $Ab$ of Abelian groups. An argument involving limits and colimits indeed can be found: finite coproducts coincide with finite products in the category of Abelian groups but they differ in the category of groups.

Notably, on the other hand, the group objects in the category of groups do form an equivalent category to $Ab$, due to the Eckmann-Hilton argument.