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

Solution 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.