Is there a such thing as a free group generated by a free group?

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.

• In the language of category theory, given a set $X$ you can construct the free group $F(X)$. And given a group $G$, you can consider its underlying set $U(G)$. You are trying to consider $F(U(F(X)))$. This is not equal to $F(X)$, but there is a natural transformation from $F\circ U\circ F$ to $F$, so that there is a nice homomorphism from $F(U(F(X)))$ to $F(X)$. Commented Mar 19, 2022 at 5:14

Let $$A=\{a\}$$. So $$F(A) \simeq \mathbb{Z}$$. Then $$F(F(A))=F(\mathbb{Z})$$ which is definitely not $$\mathbb{Z}$$. It is a free group on a countable number of generators.

The problem is when you do the second free product you get words like

$$(a \cdots a)^\pm(a \cdots a)^\pm \cdots$$. The parentheses cannot be erased. Each term in parentheses is interpreted as a single letter. When applying free group the second time (a free and a forget really), you have lost the fact that they were originally made up of $$a$$'s from $$A$$.

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.

• @ArturoMagidin I agree that was sloppy. Thanks for the correction. Commented Mar 19, 2022 at 5:34

Suppose $$A=\{x\}$$. Then $$F(A)$$ is just the set of words on $$x$$ of varying length, where two words are equal iff their lengths are equal.

Now consider $$F(F(A))$$, which looks similar, but now the words $$xxx\cdot xx\neq xxxx\cdot x$$ even if they have the "same length". In general, free groups are isomorphic iff their basis have the same cardinality.