e range $L_n[{\mathcal{F}}(G)]$ is the dimension of ${\mathcal{F}}(G_n)$, which is equal to $n$.

Since now obviously for every $x\in {\mathcal{F}}(G)$ (notice that it suffices to verify it for the dense set $\mathrm{span}\{\delta_g\colon g\in G\}$) we have $$\lim_{n\to\infty} L_n(x)=x,$$ then by [@AlbiacKalton Proposition 1.1.7] ${\mathcal{F}}(G)$ possesses a Schauder basis.

Examples {#subsec:examples}
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[*Finitely generated abelian groups*]{}. Recall that every infinite finitely generated abelian group $A$ is of the form ${\mathbb{Z}}^n\oplus F$, where $n\geq 1$ and $F=\{0,f_1,\ldots,f_m\}$ is a finite abelian group. Let $e_1,\ldots,e_n$ be the canonical generators of ${\mathbb{Z}}^n$. Then $e_1\leq -e_1\leq e_2\leq -e_2\leq\ldots\leq e_n\leq -e_n\leq f_1\leq \ldots\leq f_m$ is a linearly ordered finite symmetric generating set. We leave to the reader to verify that with this ordered generating set $A$ is shortlex combable.\
[*Free groups*]{}. Let $n\geq 1$ and let $F_n$ be a free group on $n$ generators (which is ${\mathbb{Z}}$ for $n=1$). Let $a_1,\ldots,a_n$ be the free generators. It is again an exercise that with the order $a_1\leq a^{-1}_1\leq\ldots\leq a_n\leq a_n^{-1}$, the group $F_n$ is shortlex combable. We note however, that ${\mathcal{F}}(F_n)$ is linearly isometric to $\ell_1$, thus admits monotone Schauder basis. Indeed, this can be verified directly by noticing that the set $\{\delta(g)-\delta(h)\in{\mathcal{F}}(F_n)\colon d(g,h)=1, d(g,1)>d(h,1)\}$ is equivalent to the standard basis of $\ell_1$.\
[*Hyperbolic groups*]{}. Recall that a geodesic metric space $(M,d)$ is (Rips)-hyperbolic if there exists a constant $K\geq 0$, *hyperbolicity constant*, such that for any triple of points $x,y,z\in M$ and geodesic segments $S_1,S_2,S_3$ connecting each pair of the triple we have $d_{H}(S_i,S_j\cup S_k)\leq K$, where $d_H$ is the Hausdorff distance and $i,j,k$ are pairwise different from $\{1,2,3\}$. In other words, for each $i\leq 3$ and each point $x\in S_i$, $d(x,S_j\cup S_k)\leq K$, where