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Let $F_n$ be the free group with fixed set of free generators $\{x_1,x_2,\dots,x_n\}$. Whenever $w\in F_n$, write $L(w)$ to denote the set of $i\in\{1,2,\dots,n\}$ such that $x_i$ appears as a letter in the reduced word representation of $w$ (with respect to the given generators). Fix a non-trivial cyclically reduced word $a\in F_n$ and let $w$ be an arbitrary non-trivial element of the normal subgroup $\langle\langle a\rangle\rangle$ of $F_n$ generated by $a$. Must it be that $L(a)\subseteq L(w)$?

In other words, if we consider a product of conjugates of $a$ that reduces to $w$ must some instance of each letter of $a$ survive the entire reduction process?

I expect this has a positive answer and, most likely, I'm asking something much weaker than what one can say generally about elements of $\langle\langle a\rangle\rangle$. If there is a known characterization of the reduced representatives of the elements of $\langle\langle a\rangle\rangle$ (perhaps in terms of non-crossing inverse-pairing structures) that would be even better and I'd appreciate a reference. I thought an answer might follow from free group theory developed in the 50's/60's involving Peuffer transformations. For example, this relevent paper of Cohen and Lyndon shows that $\langle\langle a\rangle\rangle$ is freely generated be a set of certain conjugates of $a$. However, this result doesn't directly answer my question and I don't see any part of the proofs within that characterize reduced representatives in the original generators. But it's possible that I'm overlooking something.

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If I understood the question correctly this is a consequence the Freiheitssatz of Magnus. This theorem says that if we take the one relator group $G=\langle x_1,\ldots,x_n\mid a\rangle$, then the subgroup of $G$ generated by any subset of $x_1,\ldots, x_n$ that does not contain all the letters of $a$ is free on that subset. In particular, if your word $w$ does not contain all the letters of $a$, then $w$ cannot be trivial in this one-relator group.

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