orders and length functions on finitely generated groups

Let $G$ be a finitely generated group with the natural word length function ($|x|$ is the length of the shortest word in generators of $G$ representing $x$). We call a partial left invariant order $\le$ on $G$ word order if whenever $a\le b\le c$ we have $|b|\le C(|a|+|c|)$ for some constant $C$. Say, the standard order on $\mathbb Z$ is a word order.

Question. Is there a group $G$ with a left invariant linear order but without left invariant linear word order?

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If $G$ is a finitely generated infinite group and $\leq$ is a linear word order, then for each $a, c \in G$ there are only finitely many elements $b \in G$ such that $a \leq b \leq c$. From this it follows that $(G, \leq)$ is order isomorphic to $\mathbb Z$. If $\leq$ is also left invariant, then this isomorphism must be a group isomorphism as well.

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@Jesse: Thanks! – Mark Sapir Oct 20 '10 at 22:08

Let $G=\mathbb Z^2$. Every invariant linear order on $\mathbb Z^2$ is either induced from the standard order on $\mathbb R$ (or its inverse) by a linear map of the form $$(x,y)\mapsto x+\alpha y : \mathbb Z^2\to\mathbb R$$ where $\alpha\in\mathbb R\setminus\mathbb Q$, or is a composition of the lexicographic order $$(x,y)>(x',y') \quad\iff\quad x>x' \text{ or } (x=x' \text{ and } y>y')$$ with a bijective linear transformation $\mathbb Z^2\to\mathbb Z^2$.

Indeed, the set of elements of $\mathbb Z^2$ that are greater or equal to zero is a semi-group $H$ such that $H\cap(-H)=\{0\}$. The closed convex hull of such $H$ must be a half plane. If this half-plane is bounded by an irrational line, we have the first type of a linear order. If this half-plane is bounded by a rational line, then we may assume that it is a coordinate line $\{x=0\}$ (up to a linear change of coordinates in $\mathbb Z^2$), then the order is the lexicographic one (up to a change $y\mapsto -y$ depending on whether $(0,1)$ is "positive" or "negative" in this ordering).

In both cases, there are infinitely many elements between $(0,0)$ and $(1,0)$, hence it is not a word order.

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@Sergei: Thanks! Jesse was 12 minutes earlier, so I will accept his answer. – Mark Sapir Oct 20 '10 at 22:10
@Mark: actually only 1 minute earlier;) His answer is so much cleaner, I would not write mine if I saw it. – Sergei Ivanov Oct 20 '10 at 22:15
@Sergei: That's correct, I looked at when it was edited, not when it was posted. It was a 1 minute difference. – Mark Sapir Oct 20 '10 at 22:31