The Cayley graphs of $\mathbb{Z}\mathrm{/}2$ for $\{1\}$ and of $\mathbb{Z}$ for $\{-1,1\}$ are trees, and neither generating set is free

Given: The proposed claim together with the witness named in the Statement refuted.

The Cayley graph of a group $G$ with respect to a subset $S$ has vertex set $G$ and edge set $\{\{g,gs\}:g\in G,s\in (S\cup S^{-1})\setminus \{e\}\}$ (The Cayley graph of a group with respect to a subset).

A cycle is a closed walk of length at least three with distinct vertices apart from its endpoints; a forest is a simple graph with no cycle and a tree is a connected forest (Cycles, trees and forests in a simple graph on an arbitrary vertex set).

If no product of two members of a generating set is the identity and the Cayley graph is a tree, the set is a free basis (If no product of two members of a generating set is the identity and the Cayley graph is a tree, the set is a free basis).

The subset $B$ is a free basis of $F$ if $(F,i)$ is a free group on the set $B$ in the sense of. (A free basis of a group).

If $G=⟨g⟩$ is cyclic, then exactly one of the following applies: (Every cyclic group is isomorphic to $(\mathbb{Z},+)$ or to $(\mathbb{Z}\mathrm{/}n,+)$ for its finite order $n\ge 1$).