It is a simple exercise to prove that if $F$ is the free group on two letters $\{x,y\}$, then the subgroup $$H = \{w \in F \mid \text{sum of all }y\text{ exponents in } w = 0 \} = \text{ normal closure of }\langle x\rangle$$ is free with the infinite basis $\{y^ixy^{-i}\mid i \in \mathbb{Z}\}$. Is there any classical/well known book that poses this (or an equivalent statement) as an exercise or a proposition? I've tried searching for it on Magnus-Karrass-Solitar but found nothing like it so far...
2026-03-25 15:54:38.1774454078
Reference for an exercise in free groups
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You can do this using covering space theory as follows. $F_2$ is the fundamental group of the wedge of two circles $X = S^1 \vee S^1$; covering spaces of $X$ correspond to subgroups of $F$ as usual. The subgroup $H$ here corresponds to the following cover $Y$: if we think of the first copy of $S^1$ as corresponding to $x$ and the second copy as corresponding to $y$, imagine "unrolling" the second copy of $S^1$ into $\mathbb{R}$, and then wedging with a copy of $S^1$ at each of the integer points $\mathbb{Z} \subset \mathbb{R}$. The covering map $Y \to X$ is given by "rolling up" $\mathbb{R}$ back into $S^1$. This would be a lot easier to explain with a diagram, which I might get around to drawing later.
The fundamental group of any graph is free on loops in bijection with the set of edges not contained in any spanning tree, and $Y$ has an obvious spanning tree given by the copy of $\mathbb{R}$ in it. There is one edge not in this spanning tree for every point in $\mathbb{Z} \subset \mathbb{R}$, and the corresponding loops in $\pi_1(Y)$ are given by going from the origin $0 \in \mathbb{R}$ to some integer point $i \in \mathbb{Z}$, then following the loop in the corresponding copy of $S^1$, then going back to the origin; this maps precisely to $y^i x y^{-i}$ as desired.
Edit: Here's a picture of $Y$ and $X$, with the edges colored to indicate the covering map. Red is $x$ and blue is $y$.