(Co)completeness of the category of abelian group objects in a (co)complete category Let $C$ be a category with all finite (co)limits and a $0$ object. Let $\mathbf{Ab}(C)$ be the category of abelian group objects in $C$, i.e. the category of representable functors $C^{op} \to Set$ that factor through the category of abelian groups. Does $\mathbf{Ab}(C)$ have all finite (co)limits? 
Equivalently, does a finite (co)limit of objects in $C$ each with an abelian group structure have a group structure? Obviously, one must assume that all the maps are group homomorphisms.
 A: For limits, yes.  The functor category $Ab^{C^{op}}$ has limits (evaluated pointwise).  Moreover, in $Set^{C^{op}}$, a finite limit of representable functors is representable, since an object representing the limit is the same thing as a limit of the representing objects, and $C$ has all finite limits.  This remains true in $Ab^{C^{op}}$ since limits are evaluated pointwise in both categories and the forgetful functor $Ab\to Set$ creates limits.   Thus $\mathbf{Ab}(C)$ has finite limits, and more generally the forgetful functor $\mathbf{Ab}(C)\to C$ creates limits.
Or if you prefer, in more concrete terms: if you take a limit of a diagram of abelian group objects, the set of maps from any other object $X$ of $C$ into the limit will be the limit of the sets of maps from $X$ to each object in the diagram (this is basically the definition of a limit).  Since the limit (in $Set$) of a diagram of abelian groups and homomorphisms has a natural abelian group structure, this means the limit object is an abelian group object.
For colimits the story is not so nice.  We see this already in the case $C=Set$, where the forgetful functor $Ab\to Set$ fails to preserve colimits (although colimits do exist in $Ab$).  For an example where finite colimits exist in $C$ but not in $\mathbf{Ab}(C)$, let $C$ be the poset $\mathbb{N}$.  Then $\mathbf{Ab}(C)$ is the empty category, since if $X$ is an abelian group object then $\operatorname{Hom}_C(Y,X)$ is an abelian group and in particular is nonempty for all $Y$, but there is no object of $C$ that has a map from every other object.  In particular, $\mathbf{Ab}(C)$ does not have an initial object.
(However, if $C$ has finite limits, then $\mathbf{Ab}(C)$ at least has finite coproducts, since finite products and finite coproducts are the same in $\mathbf{Ab}(C)$ and more generally in any category enriched in abelian groups.  I don't know of an example where $C$ has both finite colimits and finite limits but $\mathbf{Ab}(C)$ fails to have coequalizers, but I think that should be possible.)
A: Another way to approach the case of finite limits existing is as follows.
Abelian group objects in $C$ are models of a finite limit sketch.
If you're not familiar with the term, the least technical description is to let $\mathbf{T}$ be the opposite category of the category of finitely presented groups. Then an abelian group object in $C$ is the same thing as a functor $M : \mathbf{T} \to C$ that is left exact — i.e. preserves finite limits.
(The underlying object is $M(\mathbb{Z})$. Addition is the map $M(\mathbb{Z}^2) \cong M(\mathbb{Z})^2 \to M(\mathbb{Z})$ induced by the diagonal $\mathbb{Z} \to \mathbb{Z}^2$. And similarly for the other operations)
Since $C$ has finite limits, finite limits in the functor category $C^{\mathbf{T}}$ are computed pointwise. And we can see that a limit of left exact functors is left exact by the following argument.
Let $F = \lim_i F_i$ be a finite limit of functors and $t = \lim_j t_j$ be a finite limit of objects of $\mathbf{T}$. Assuming the $F_i$ preserve finite limits, interchange of limits proves that $F$ does as well:
$$\begin{align} F(t) &= (\lim_i F_i)(t)
= \lim_i F_i(t)
\\&= \lim_i \left( F_i(\lim_j t_j) \right)
= \lim_i \left( \lim_j F_i(t_j) \right)
\\&= \lim_j \left( \lim_i F_i(t_j) \right)
= \lim_j \left( (\lim_i F_i)(t_j) \right)
\\&= \lim_j F(t_j)
\end{align}$$
Note that if $C$ has all small limits, then the same argument applies to show $\mathbf{Ab}(C)$ has all small limits.
Also of note is that if $C$ has filtered colimits that commute with finite limits, then $\mathbf{Ab}(C)$ also has filtered colimits that commute with finite limits.
