Grp is complete and cocomplete

For groups $(G_i{)}_{i\in I}$, the Cartesian product with componentwise multiplication and inverse is a group. Coordinate projections are homomorphisms, and the unique set-theoretic pairing of a family of homomorphisms is componentwise a homomorphism. This gives every small product, including the one-element group for $I=\mathrm{\varnothing }$.

For $f,g:G\rightrightarrows H$, the subset $E=\{x\in G:f(x)=g(x)\}$ is a subgroup. Its inclusion is an equalizer because an equalizing homomorphism has image in $E$ and corestricts uniquely. Thus [L1] gives completeness.

For a family $(G_i)$, let $S={\coprod }_i\{i\}\times \mathrm{\mid }{G}_i\mathrm{\mid }$, take the free group $F(S)$, and quotient by the normal closure $N$ of the elements imposing $[i,1]=1$ and $[i,xy]=[i,x][i,y]$. The maps $G_i\to F(S)\mathrm{/}N$, $x\mapsto [i,x]N$, are homomorphisms.

For $f,g:G\rightrightarrows H$, let $N$ be the normal closure of $\{f(x)g(x{)}^{-1}:x\in G\}$. The quotient $q:H\to H\mathrm{/}N$ satisfies $qf=qg$. If $h:H\to K$ equalizes $f,g$, it kills those generators and their normal closure, so [L2] gives a unique factor through $q$. Thus $q$ is a coequalizer.

Given homomorphisms $h_i:G_i\to H$, [F2] extends the tagged function to a unique $F(S)\to H$. It kills every defining relation, hence $N$, so [L2] factors it uniquely through $F(S)\mathrm{/}N$. Conversely any factor restricts to the $h_i$. This proves the coproduct existence and uniqueness clauses, including the empty family, where the free group is trivial.

Steps 1.3, 2.1, and 1.4 give all small coproducts and coequalizers. The dual half of [L1] therefore gives cocompleteness.