Word length of a group element with respect to a generating set

Let $G$ be a group and let $S\subseteq G$ be a generating set. For $g\in G$, consider the set

$L_S(g):=\{n\in \mathbb{N}:g=s_1\cdots s_n\text{ for some }{s}_1,\dots ,s_n\in S\cup S^{-1}\}\mathrm{.}$

This set is nonempty because $S$ generates $G$. Indeed, the set of all finite products of elements of $S\cup S^{-1}$ contains the identity, is closed under products and inverses, and contains $S$, so it is a subgroup containing $S$; conversely every subgroup containing $S$ contains all such products. It is therefore exactly $⟨S⟩=G$ by The subgroup $⟨S⟩$ generated by a subset, the cyclic subgroup $⟨g⟩$, and cyclic groups. By The well-ordering principle, it has a least element. The word length of $g$ with respect to $S$ is that least element and is written

$\mathrm{\mid }g{\mathrm{\mid }}_S:=\min L_S(g)\mathrm{.}$

Thus $\mathrm{\mid }g{\mathrm{\mid }}_S=0$ exactly when $g$ is the empty product, that is, the identity of $G$. When $S$ is finite, such expressions are obtained by evaluating words in the formal alphabet $S\bigsqcup S^{-1}$ of Words in an alphabet with formal inverses, elementary cancellation, and reduced words; distinct formal words can evaluate to the same product when two letters represent the same group element of $G$.