TITLE: How to recover multiplication of group elements from category of groups?
QUESTION [8 upvotes]: Motivating question: If we know everything about $\mathbf{Grp}$, do we know everything about groups? 
Background: Consider the category $\mathbf{Grp}$ abstractly, rather than as a concrete category.  The elements of a group $G$ may be recovered as follows: first, identify the group $\mathcal{Z}$ as the unique minimal generator (MO/7793), i.e. a generator such that no proper quotient is also a generator.    Then Hom($\mathbb{Z}$,-) represents the forgetful functor $\mathbf{Grp} \to \mathbf{Set}$ and recovers the elements of group.
Question:

How can you also recover the group operation on elements?  

Here is an unfinished attempt (that someone else mentioned): define the free group on two generators by the coproduct $F_2 = \mathbb{Z} \amalg \mathbb{Z}$.  Then for two group elements $g_1, g_2 : \mathbb{Z} \to G$, these maps will define a map $f :  F_2 \to G$ (because $F_2$ is a colimit).  Finally, suppose we could categorically specify the map $m : \mathbb{Z} \to F_2, 1 \mapsto xy$, where $x,y$ are the generators of $F_2$.  Then the group multiplication of the elements $g_1, g_2$ would yield the element $g_1 g_2 = f \cdot m : \mathbb{Z} \to G$.
In the above approach, the only remaining task is to somehow specify the map $m : \mathbb{Z} \to F_2, 1 \mapsto xy$ categorically.
Edit (in response to comment):
I should mention that we will not be able to distinguish $m : \mathbb{Z} \to F_2, 1 \mapsto xy$ from $m' : 1 \mapsto yx$.  The best we can hope for is to specify both $m$ and $m'$ without singling either of them out.  In other words, when we attempt to recover group multiplication, we can provide a choice of either the original group multiplication or multiplication in the opposite group, but we cannot privilege either choice.  The reason is that the functor sending a group to its opposite group is an auto-equivalence of $\mathbf{Grp}$.

REPLY [2 votes]: Once you have $\mathbb{Z}$, you have the Lawvere theory of groups (which can be recovered as the opposite category on the full subcategory on the finite coproducts of $\mathbb{Z}$), which is a way of axiomatizing what a group is that doesn't depend on a choice of "generators." What I mean by this is that multiplication, inversion, and unit can be composed to get a lot of other operations, such as
$$(x, y, z) \mapsto x^{-3} y z^4 y^3$$
which satisfy some relations, and the Lawvere theory of groups packages up all of this information (the above operation is, after all, the same thing as a particular element of $F_3$) without privileging a particular choice of generating operations, in the same way that you can specify a group itself in a way that doesn't privilege a particular presentation. See, for example, this blog post and this blog post for some details.