TITLE: Are semi-direct products categorical (co)limits?
QUESTION [31 upvotes]: Products, are very elementary forms of categorical limits. My question is whether in the category of groups, semi-direct products are categorical limits.
As was pointed in:
http://unapologetic.wordpress.com/2007/03/08/split-exact-sequences-and-semidirect-products/
Bourbaki (General Topology, Prop. 27) gives a universal property:
Let $f \colon N \to G$, $g \colon H \to G$ be two homomorphisms into a group $G$, such that $f(\phi_h(n)) = g(h)f(n)g(h^{-1})$ for all $n \in N$, $h \in H$. Then there is a unique homomorphism $k \colon N \rtimes H \to G$ extending $f$ and $g$ in the usual sense.
However, I remain unsatisfied. The condition $f(\phi_h(n)) = g(h)f(n)g(h^{-1})$ is a condition on elements of groups, rather than a condition that says that some diagram is commutative.
So the question remains: are semi-direct products in the category of groups categorical limits?

REPLY [50 votes]: There is (another ?) description of the crossed product in categorical terms. 
Let ${\rm Mor}(Gp)$ be the category whose objects are homomorphisms of groups and morphisms are commutative diagrams. Let $C$ be the category of "groups acting on groups" whose objects are pairs of groups $(H,G)$ together with a homomorphism $H \to {\rm Aut}(G)$. Morphisms in this category are equivariant homomorphisms.
Now, there is a natural forgetful functor $T \colon {\rm Mor}(Gp) \to C$ which sends $H \to G$ to the pair $(H,G)$ with the homomorphism $H \to {\rm Aut}(G)$ given by conjugation. Now, almost by definition, the crossed product is the left-adjoint of this forgetful functor. Indeed, the left adjoint is easily seen to map $(H,G)$ with $H \to {\rm Aut}(G)$ to the inclusion $H \to G \rtimes H$.
Being a left-adjoint, the "crossed product" maps colimits to colimits.