I have a doubt about something: if $G$ is a group, what can we say about the free group ${\rm Free}(U(G))$ of $U(G)$, where $U$ is the forgetful functor from $Grp$ to $Set$?
I'm confused because if I'm not wrong: every group is a quotient of a free group by some relations, so when one "forgets" about the group structure of $G$:
do we forget about these relations (in which case one would not recover $G$ by taking the free group over $U(G)$ in general)
or are these relations actual equalities in the carrier set (e.g. if we have the group relation $ab=c$, then $ab$ and $c$ are really the same element in the carrier set), in which case I have the impression that ${\rm Free}(U(G))$ is isomorphic to $G$?
I really feel like there's something wrong:
on the one hand, I suppose that ${\rm Free}(U(G))$ shouldn't be isomorphic to $G$ in general (and that to recover $G$ from ${\rm Free}(U(G))$ one would need to "evaluate" the lists of elements of $U(G)$ that are in ${\rm Free}(U(G))$ with the group operation of $G$)
but on the other hand, I feel like the group relations are actual equalities in the carrier set, so that taking the free group over this carrier set should yield a group isomorphic to the original one
Any help would be very much appreciated!
No, $\mathrm{Free}(U(G)) \neq G$. For example, if $G$ is the trivial group, then $U(G)$ is a one element set, so $\mathrm{Free}(U(G)) \cong \mathbb Z.$ In general, $\mathrm{Free}(U(G))$ is much larger than $G$.
Another example. Let $G=\mathbb Z/2\mathbb Z=\{a,b\}$, equipped with a binary operation $ab=ba=b$, $aa=bb=a$. Then $U(G)=\{0,1\}$, just as a set. In $\mathrm{Free}(U(G))$, the same multiplication structure does not hold. Every element of $\mathrm{Free}(U(G))$ is a string of symbols taken from the list $$ a,a^{-1},b,b^{-1}. $$ where the only relations that hold are $aa^{-1}$ and $bb^{-1}$ are equal to the identity.