Let $G$ be a finitely generated group, S is a set of generators. If $\forall s\in S, n\in \mathbb{Z}$, $\exists C>0$ such that $s^n_S\ge Cn$, does it imply $\forall$ infinite order element $g\in G$, $n\in \mathbb{Z}$, $\exists C>0$ such that $g^n_S\ge Cn$? If not, is there any condition to ensure it? Thanks in advance.

2$\begingroup$ This sounds very unlikely. If $G=G_1\times G_2$ where $G_2$ does not have this property, then you can choose generators that all have nontrivial parts in $G_1$ and $G_2$. But if $g$ is from the $G_2$ part, it will fail to have your property, right? $\endgroup$ – Anthony Quas May 26 '14 at 20:22
From the way you written it, it looks as though $C$ is allowed to depend on both $g$ and $n$, in which case the statement is obviously true with $C = g^n_S/n$ (or $0$ when $n=0$), but I suspect that you don't want $C$ to depend on $n$.
Let $G$ be the BaumslagSolitar group ${\rm BS}(1,2) = \langle x,y \mid y^{1}xy=x^2 \rangle$, and $S=\{ y,xy \}$.
Then, since $G/[G,G]$ is infinite cyclic and generated by the image of $y$, $y^n_S = (xy)^n_S = n$, but $x^{2^n} = y^{n}(xy)(y^{1})y^n$, so $x^{2^n}_S \le 2n$.
Another familiar example is the Heisenberg group $\langle x,y \mid [[x,y],x]=[[x,y],y]=1 \rangle$, with $S=\{x,y\}$. Then $x^n_S = y^n_S=n$, but $[x,y]^{n^2}=[x^n,y^n]$, so $[x,y]^{n^2}_S \le 4n$.