Is it true that $\leq $ is independence of generating set?

Question

Let $G$ be a finitely generated group and let $S$ be a finite generating set. (For convenience, we always assume our generating set is symmetric, i.e. $s\in S$ implies that $s^{-1}\in S$.) We define the word length $||g||_S$ of an element $g\in G- \{e\}$, with respect to the generators $S$, by $ ||g||_S=\inf\{k\geq 1: g= s_k s_{k-1}\ldots s_1, s_i\in S, 1\leq i\leq k\} $.

For $g, h\in G$, $h\leq_S g$ if and only if $||h||_S\leq ||g||_S$.

Question: Let $S,S^{*}$ be finite generating sets for $G$. Is it true that if $g\leq_S h$, then $g\leq_{S^{*}}h$?

A relation $\geq$ directs a nonempty set $D$ if an only if following are satisfied:

1. $\leq $ is reflexive and transitive

2. For each $a,b \in D$, there is a $c \in D$ such that $c \geq a$ and $c \geq b$

The pair $(D, \geq)$ is called a directed set.

It can be seen that $(G, \leq_S)$ is a directed set.

Answer 1

It is not true that word length comparisons are preserved by different generating sets. Just add whatever element you are comparing to a generating set, to make it length $1$.

Another simple, concrete example, consider the intergers $\mathbb Z$ under addition, and the standard generating set is $\{1, -1\}$, and you could also consider the generating set $\{\pm 2, \pm 3\}$, in one $2$ is longer than $1$ and in the other $1$ is longer than $2$. You can get more involved with different examples, like for non-abelian free groups you can consider different basis, and compare lengths between them.

There are restrictions though on how the different $\leq_X$ can be though, and in some sense this restriction is a starting point for the field geometric group theory. It ends up being that the "large scale geometry" does not change in a finitely generated group with different finite generating sets, and this is encoded by the concept of quasi-isometry. If $S,T$ are finite generating sets then it turns out that there is a constant $C$ such that $$\frac{1}{C} d_S(x,y)-C \leq d_T(x,y) \leq C d_S(x,y)+C,$$ where $d_X(x,y)$ is the length of $xy^{-1}$ with respect to the generating set $X$ (this can be thought of as the distance in the Cayley graph too). Drawing a more direct connection to your question, this means that $$\frac{1}{C} ||g||_S-C \leq ||g||_T \leq C ||g||_S+C.$$ Note that $C$ does depend on $S$ and $T$. Basically this says the global geometry is not that different.