Given (possibly non-Abelian) groups $H,G$ with $H \subseteq G$ and $f \in G$, I write $\langle H, f \rangle$ for the subgroup of $G$ generated by $H \cup \{f\}$.


Write $T(H)$ for the free product of $H$ with a multiplicative copy $y^{\mathbb{Z}}$ of the group of integers. We can represent a generic element $t(y)$ of $T(H)$ as a "term" $t(y) = h\_0 y^{\alpha\_0} \cdot \cdot \cdot h\_n y^{\alpha\_n}$ where $n \in \mathbb{N}$, $h\_0,...,h\_n \in H$ and $\alpha\_0,...,\alpha\_n \in \mathbb{Z}$. We set $t(f) := h\_0 f^{\alpha\_0} \cdot \cdot \cdot h\_n f^{\alpha\_n} \in G$ for this term.


Say that $f$ *cancels a term over $H$* if there is a $t(y) \in T(H) \setminus \{1\}$ with $t(f)=1$. This is the case for instance if $f$ commutes with a non-trivial element of $H$.


**Question**: Is it the case that if $f$ cancels a term over $H$, then each element of $\langle H, f\rangle$ cancels a term over $H$?


I suspect not, but I don't have a way to produce a counterexample.



Elements in group extensions which cancel unary terms in the language of groups
No. Let $G = \langle x, y \rangle$ be a $2$-generated non-free group with $x$ of infinite order such that $F = \langle x, g\rangle \cong F\_2$ for some $g \in G$. Then we get a counterexample by taking $H = \langle x \rangle$ and $f = y$. Indeed, since $G$ is not free, there is some nontrivial word $w \in F\_2$ such that $w(x, y) = 1$, so "$y$ cancels a term over $\langle x \rangle$", but since $\langle x, g\rangle$ is free the element $g$ does not cancel a term over $\langle x \rangle$.


There are probably many constructions of such a $G$. Here is one. Start with $F = F\_2 = \langle x, g\rangle$. Let $y$ be the automorphism of $F$ that swaps $x$ and $g$ and let $G = F\langle y \rangle \cong F \rtimes C\_2$. Observe that $G = \langle x, y\rangle$, since $g = x^y$. (In this case $y$ cancels a term over $\langle x \rangle$ for the boring reason that $y^2 = 1$, but $\langle x, x^y\rangle$ is free.)


