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Is there any length function on additive group of $\mathbb{Q}$ such that $\mathbb{Q}$ is of polynomial growth WRT this length function? What about the multiplicative group of $\mathbb{Q}$ instead?

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Is $\mathbb{Q} =\mathcal{Q}$ the rationals? –  plusepsilon.de May 21 '12 at 17:34
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Take the length function defined by the generating set which is the entire group. –  Lee Mosher May 21 '12 at 17:41
    
sorry my mistake I meant $\mathbb{Q}$. If I take the length function defined by a generating set its growth would be infinite. I need somehow manipulate the length function coming from a generating set to get a polynomial growth. And if it is impossible I need to prove it. –  Vahid Shirbisheh May 21 '12 at 18:01
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@Vahid: What is your definition of a length function? The standard notion of word length is used only for finitely-generated groups. I guess, you want to define $|x|$ axiomatically so that $|x+y|\le |x|+|y|$, $\{x: |x|\le n\}$ is finite for all $n\in {\mathbb N}$, etc. Until you clearly state what axioms you want, your question makes no sense. –  Misha May 21 '12 at 18:36
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@Misha. You're right; actually for a f.g. group, the word length achieves the minimal growth rate among all lengths. The interest is to make things meaningful with infinitely generated groups. In this case there is no "largest" length and, in general, no length with "smallest " growth. Lengths are also useful when you want to consider countable groups as metric spaces up to coarse equivalence (extending viewing f.g. groups as metric spaces up to QI). [I refer to Lyndon's lengths as "Lyndon lengths" because it's a much more specific context in my opinion.] –  YCor May 21 '12 at 21:39
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Yes for $\mathbf{Q}$, no for $\mathbf{Q}^*$.

For $\mathbf{Q}$, write it as the union of an increasing sequence $L_n$ with $L_1=\mathbf{Z}$ and $L_n$ of finite index over $L_1$. Pick a function $F$ with fast growth and define $l'(r)=|r|+F(\sup\{n:r\notin L_n\})$.

For $\mathbf{Q}^*$, it contains a subgroup isomorphic to $\mathbf{Z}^d$ for every $d$, so for every length the growth is at least polynomial of degree $d$. So it can't be polynomial.

Added-1: for $\mathbf{Q}$ we can arrange so that $[L_n:L_1]\le n$ for all $n$. Then we can pick $F$ to be the identity and then the growth is at most quadratic.

Added-2: the argument extends, showing that an abelian group admits a length with polynomial growth iff it's countable and has finite $\mathbf{Q}$-rank.

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$F$ does not have to be tremendously fast-growing. $F=[L_n:L_0]$ will make it a group of quadratic growth. –  Will Sawin May 21 '12 at 20:46
    
@Will I was unable to check your claim (which looks correct) but I edited taking it into account. @Lee: I maintain $\sup$ and $\notin$ (but fixed a confusing typo since I forgot the word "union"). –  YCor May 21 '12 at 22:17
    
@Yves Thank you very much for your answers. @Will Thank you for your comment. Please see my other question titled "property (RD)" for $\mathbb{Q}$ where I am going to answer it using Yves' help (in the next 30 minuets). –  Vahid Shirbisheh May 22 '12 at 3:25
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