TITLE: Is there a such thing as a free group generated by a free group?
QUESTION [6 upvotes]: Is there a such thing as a free group generated by a free group?
Let $F(A)$ be a free group generated by the elements of a set $A$. If we momentarily consider $F(A)$ to be simply a set of words on elements of $A$, is there a such thing as $F(F(A))$, i.e., the free group generated by the elements of $F(A)$?
My guess is that this free group would still just be $F(A)$?
Since it will be all possible words on elements of $F(A)$, which are also words on $A$, therefore the elements of $F(F(A))$ will simply be products of words on $A$, but those would already be in $F(A)$ by definition of a free group, so they would be the same.

REPLY [8 votes]: You're essentially asking whether the free group construction is idempotent. This

Since it will be all possible words on elements of $F(A)$, which are also words on $A$,

is where you go wrong.
Let $A = \{a, b, c, d, e\}$. Then
\begin{align*}
ab^3a^{-1}c &\in F(A)\\
d^{-2}a^2be^{-1}c^3a^{-1} &\in F(A).
\end{align*}
Now consider $F\big(F(A)\big)$. We have
\begin{align*}
d^{-2}a^2be^{-1}c^3a^{-1} \cdot ab^3a^{-1}c \in F\big(F(A)\big)
\end{align*}
You can't just collapse that $a^{-1} \cdot a$ in the middle because then you're switching definitions mid-stream. You can't start breaking apart the elements of $F(A)$ like that.