If $G$ is finitely generating group with generating set $S=\lbrace s_1,..., s_n\rbrace$. Then we need $S$ to be a symmetric set in order to define the word metric.
Is it possible that, if $G$ is abelian than condition on $S$ to be symmetric, can be drop? As in case of integers, group is cyclic but we can still define the length. Is it possible, when $G$ is abelian than we can simply take the length of $g\in G$ to be the $infimum$ of $sum$ of modulus of $powers$ of $s_i's$ where $i\in \lbrace 1,..., n\rbrace$ and $infimum$ is taken over distinct representation's of $g$.
Mainly the problem is: Let $S=\lbrace s_1,..., s_k\rbrace$ (need not symmetric set i.e we do not need $a\in S$ if and only if $a^{-1}\in S$) be a finite generating set for abelian group $G$. Then every element of $g\in G$ can be written as a representation $R_g=s_{1}^{n_1}...s_{k}^{n_k}$ where $n_{i} \in Z$. Define $R^{+} _{g} =|n_1|+...+ |n_k|$ and length of $g$ to be$L_g= inf_{R_g} R^{+}_{g} $. Clearly, infimum exist. Define metric as $d(g, h) =L_{g^{-1}h}$ I am claiming that for such definition of word length on finitely generated abelian group, $d$ defines a metric.