TITLE: Abstract nonsense proof of the cocompleteness of the category of groups
QUESTION [8 upvotes]: There is a well-known way to conclude the cocompleteness of a category from its completeness. Namely, if a category is complete, well-powered and has a cogenerating set, then it is cocomplete (easy corollary of the special adjoint functor theorem).
I was hoping I could use this to prove that the category $\mathbf{Grp}$ of groups is cocomplete. I would find it interesting, for completeness of $\mathbf{Grp}$ is easy to prove (the forgetful functor to the category of sets creates limits), while cocompletenes is not so easy (constructing free product is quite tedious).
But alas, a quick google search revealed that $\mathbf{Grp}$ doesn't have a cogenerator (it can be read off the only freely available page of this article). I suspect it doesn't have a cogenerating set, either.
Is there a categorical proof of the cocompleteness of $\mathbf{Grp}$?
Edit: I'd like to point out that I found out afterwards the following exercise 1 in section IX.1 of Mac Lane's CWM (I didn't expect to find this exercise in the chapter on "Special Limits"!)

Use the adjoint functor theorem to prove in one step that Grp has all small colimits.

REPLY [2 votes]: Zhen Lin already sketched a proof of the cocompleteness of $\mathbf{Grp}$ using the general adjoint functor theorem. Here's how I filled up the details.
Let $J$ be a small category; we wish to show the diagonal functor $\Delta:\mathbf{Grp}\to \mathbf{Grp}^J$ has a left adjoint. The diagonal functor preserves limits (direct consequence of the pointwise construction of limits in $\mathbf{Grp}^J$) and $\mathbf{Grp}$ is complete, so it suffices to check the solution set condition.
Let $D\in \mathbf{Grp}^J$ be a diagram. We note $D_j:=Dj$ for all $j\in J$.
Consider the class consisting of groups of cardinality $\aleph_0$ and of groups of cardinality $\leq \lvert \bigsqcup\limits_{j\in J} D_j \rvert$ for all $j\in J$. We consider the set $\{G_\alpha\}_{\alpha\in \Lambda}$ of groups consisting of one isomorphism class per each of these groups.
I claim the set of all natural transformations $\{D\Rightarrow \Delta G_\alpha\}_{\alpha\in \Lambda}$ is a solution set.
Fix $\varphi:D\Rightarrow \Delta G$. We must check that there exists $\alpha\in \Lambda$, $\tau^\alpha:D\Rightarrow \Delta G_\alpha$ in the solution set and $t:G_\alpha\to G$ a group homomorphism such that the following diagram commutes:

Remember that natural transformations $D\Rightarrow \Delta K$ are exactly cocones for the diagram $D$ with vertex $K$.
So, to put it differently, we must find a cocone morphism between some cocone $\tau^\alpha$ in the solution set and the fixed cocone $\varphi$, i.e. a group morphism $t:G\alpha\to G$ such that the following diagram commutes for every arrow $e^i_j:i\to j$ of $J$:

Let $H$ be the subgroup of $G$ generated by $\bigsqcup\limits_{i\in J} \operatorname{Im} \varphi_j$. Then $H=G_\alpha$ for some $\alpha\in \Lambda$ (by the same argument used in the proof that there exist free groups using GAFT).
We define $\tau^\alpha_i:D_i\to G_\alpha$ as $\tau^\alpha_i(x)=(\varphi_i(x),i)$ for every $i\in J$. So $\tau^\alpha:D\Rightarrow \Delta G_\alpha$ is a cocone for $D$: this is quickly verified using the fact that $\varphi$ is already a cocone.
Now we define $t:G_\alpha\to G$ as $t(\varphi_i(x),i)=\varphi_i(x)$; clearly $t\circ \tau^\alpha_i=\varphi_i$ for all $i\in J$, thus finishing the proof.