Every group is an F-algebra, but is every F-algebra (of the right signature) a group object? Wikipedia's F-algebra page explains how, given
(1) a category $C$ of group objects with finite products / coproducts and a terminal object 1, and
(2) a group object $G \in Ob(C)$,
we can define an F-algebra $(G,\alpha)$ for the functor $F(X) = 1 + X + X \times X$ where $\alpha : 1 + G + G \times G \rightarrow G$ is defined by sending any elements of 1 to the identity of $G$ in the first part of the coproduct, elements of G are sent to inverses in the second part, and we have group multiplication in the third part.
Question:  Wikipedia actually leaves off the requirement that every object in $C$ be a group object, but I added that requirement to (1) because I don't know how to define $\alpha$ if I don't know a priori of inverse or multiplication maps.  Was I correct in assuming this added assumption is needed or is there a way to construct $\alpha$ even if I don't know that all objects are group objects beforehand?
Thanks so much for any input! :)
 A: I think you are misunderstanding the construction slightly. The construction is this: let $C$ be a category with finite products and coproducts (for me "finite" implies the empty case so this implies $C$ has an initial and terminal object), which in Wikipedia's example is just $\text{Set}$. Let $G$ be a group object in $C$. Then $G$ determines an $F$-algebra for the functor $F(X) = 1 + X + X^2$, on $C$ (not on the category of group objects in $C$), where the three components of the map
$$(1 + G + G^2) \to G$$
are, respectively, the inclusion of the identity, the inverse map, and the multiplication map. This map is a morphism in $C$, not in the category of group objects in $C$, because the inverse and multiplication aren't group homomorphisms in general (in fact this is equivalent to the group being abelian).
It doesn't matter whether you know that any other object of $C$ is a group object; the construction is about a particular group object in $C$, namely $G$.
As for the question in your title, which you never ask in the body, no, being an $F$-algebra just means being an object equipped with three maps $1 \to G, G \to G, G^2 \to G$ and no axioms imposed on those maps. There is a version of this construction that is capable of imposing the group axioms and it involves taking $F$ to be a monad rather than just a functor.
