Why is the category of groups not closed, or enriched over itself? Wikipedia has some cryptic things to say about the non-abelian structure of the category of groups.
It says the category of groups is not additive because "there is no natural way to define the sum of group homomorphisms". What about just adding them pointwise? I understand groups are not additive because they're not $\mathsf{Ab}$-enriched, so pointwise addition would be commutative. But then, aren't groups enriched over themselves at least? I remember reading about a non-abelian tensor product somewhere...
Even if not enriched over itself, is it not at least closed?
 A: It's a consequence of the famous Eckmann-Hilton argument. Let $u,v : G \to H$ be two groups homomorphisms. Define $w : G \to H$ be $w(g) = u(g) v(g)$ as you suggest. Then for $w$ to be a group homomorphism, you need:
$$w(gh) = u(gh) v(gh) = u(g) u(h) v(g) v(h) \\
= w(g) w(h) = u(g) v(g) u(h) v(h) \\
\implies u(h) v(g) = v(g) u(h)$$
and so this defines a group homomorphism iff $u(h)$ commutes with $v(g)$ for all $h,g \in G$. Obviously, this is not necessarily the case, so in general pointwise multiplication doesn't yield a group homomorphism. It's not that "addition" of group homomorphisms isn't commutative that prevents $\mathsf{Grp}$ from being abelian; it's that it isn't even defined.
A lot of work has been done to try and see what properties characterize $\mathsf{Grp}$; as far as I know, semi-abelian categories capture a great deal about $\mathsf{Grp}$.
A: The category of groups is not additive because products and coproducts disagree. If they agreed, you would not only have a natural monoid structure on homsets, it would necessarily be commutative; see this blog post for details. 
However, it is true that the category of groupoids is cartesian closed; the category of functors between two groupoids is naturally a groupoid itself. For two groups this works out to the following: $\text{Hom}(G, H)$ is the groupoid whose objects are homomorphisms $f : G \to H$ and where morphisms are given by pointwise conjugating with elements of $H$. 
Edit: Also, the category of groups is not (cartesian) closed because the functor $(-) \times G$ does not preserve colimits. In fact it already does not preserve coproducts: $(\mathbb{Z} \sqcup \mathbb{Z}) \times \mathbb{Z}$ is not isomorphic to $(\mathbb{Z} \times \mathbb{Z}) \sqcup (\mathbb{Z} \times \mathbb{Z})$. The nonabelian tensor product doesn't do anything interesting to just a pair of groups (with no actions specified): you just get the tensor product of their abelianizations. This isn't a monoidal structure on groups because it has no unit. 
