Is the free group functor an isomorphism? When I have a group $G$, how come the free group functor $F: \mathbf{Set} \to \mathbf{Grp}$ is not an isomorphism with the forgetful functor $G: \mathbf{Grp} \to \mathbf{Set}$?
It seems like for any set, I might as well just add the identity element, the inverse elements, and use string concatenation as the multiply.
However I usually hear that the free functor is left adjoint to the forgetful functor, $F \dashv G$, however it seems to that I might as well turn this around?
 A: OK, so you can do that to convert a set to a group. 
Now take that group and apply the forgetful functor: what set do you get? 
In particular, if $S = \{a\}$, then $G$ looks like the group $\Bbb Z$. Now forget the group structure and you've got an infinite set, not a singleton set!
A: The forgetful functor you mention does not preserve limits, so it is not an iso. Moreover, the two categories are not isomorphic, nor even equivalent, for instance because subobjects behave very differently.
A: *

*One question is whether the free (F) and forgetful (G) functors on Grp and Set are inverses of each other. It turns out that they are not inverses of each other. One way to see this is that the free functor (F) always increases the number of elements you have: the group of "words on X" contains all the elements of X as "singleton words", plus inverses, unit, and all possible concatenations of these. In contrast, the forgetful functor preserves the number of elements you have. For this reason, $F\circ G$ and $G\circ F$ are not identity-like.

*Second, there's the question of whether in addition to $F\vdash G$, we also have $G\vdash F$.
The free (F) and forgetful (G) functors on Grp and Set are adjoints of each other. According to one definition, this means that we have that there's a family of natural isomorphisms:
$$\hom_\mathbf{Grp}(FX, Y) \cong \hom_\mathbf{Set}(X, GY)$$
because a $\mathbf{Set}$ function between a set $X$ and the elements of a group $Y$ uniquely and characteristically correspond to a morphism between the free group of "words" on $X$ and the group $Y$ itself. Specifically, the mapping sends the $\mathbf{Set}$ function $f : X\rightarrow GY$ to the function on $\mathbf{Grp}$  which sends $x_1\ldots x_n$ to $f(x_1)\oplus f(x_2)\oplus \ldots \oplus f(x_n)$. (Conversely, any group morphism sending the free group of words on $X$ to $Y$ characteristically defines a function from the underlying set $X$; the function is determined by the effect of the morphism on singleton words.)
Are the functors adjoint the other way, where $G\vdash F$? This would mean we have a family of natural isomorphisms 
$$\hom_\mathbf{Set}(GX, Y) \cong \hom_\mathbf{Grp}(X, FY)$$
but in fact sometimes these hom sets aren't isomorphic at all, let alone naturally!  Hopefully it will not be confusing if we choose the following example:
$$X = \langle \mathbb{Z}, + \rangle\qquad Y=\{1\}\\GX = \mathbb{Z}\qquad\qquad FY=\langle \mathbb{Z}, +\rangle$$
There is exactly one morphism in $\hom_\mathbf{Set}(GX, Y) = \hom(\mathbb{Z}, \{1\})$, namely the constant function sending every member of $\mathbb{Z}$ to the single point in $Y$. On the other hand, there are a lot of morphisms in $\hom_\textbf{Grp}(X, FY) = \hom(\langle \mathbb{Z}, +\rangle, \langle \mathbb{Z}, +\rangle)$. In particular, for each multiple $a$, we can define a unique morphism $n \mapsto an$ — there are infinitely many such morphisms. 
So the sets $\hom_\textbf{Grp}(FX, Y)$ and $\hom_{\textbf{Set}}(X, GY)$ are sometimes not isomorphic at all, let alone naturally in the way required for an adjunction.  Hence although $F \vdash G$,  it's not true that $G \vdash F$.
