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.

  • 3
    $\begingroup$ 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 $F(X)$. $\endgroup$ Mar 19, 2022 at 5:14

3 Answers 3


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 When applying free group the second time (a free and a forget really), you have lost the fact that they were originally made up of $a$'s from $A$.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.

  • $\begingroup$ @ArturoMagidin I agree that was sloppy. Thanks for the correction. $\endgroup$
    – Novice
    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.


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