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Suppose that we are given some finitely generated group $ G $ and some finite index subgroup of it $ H $. Given a finite generating symmetric generating set $ S \subset G $, we can define the word metric on $ G $:

$$ d_{S}(g,x):= |g^{-1}x|=min(n\in \mathbb{N} : \exists s_{1},...,s_{n}\in S, g^{-1}h=s_{1}\cdot s_{2}\cdot \dots\cdot s_{n}=g^{-1}x) $$ for all $ g,x \in G $

Similarly, take some symmetric generating set $ T \subset H $ and consider the word metric $ d_{T}^{\prime} $ on $H$. It is well known that these are quasi-isometric on H - for some constants $ K>0 , C>0$, we have:

$$ \frac{1}{K} d_S(h,y)-C \le d^{\prime}_{T}(h,y) \le K d_S(h,y)+C \;\;\forall h,y\in H $$

This property is used many times in geometric problems, and it is quite useful.

However, i'm studying the horofunction boundary (especially for nilpotent groups), in which expressions of the form $ d_S(g,x)-d_S(g,x_{0}) $ appear all the time. I want to know how to relate these to expressions of the form $ d_{T}^{\prime}(g,x)-d_{T}^{\prime}(g,x_{0}) $.

This motivates the following questions:

  1. Given a symmetric generating set $ S \subset G $, can one always find a symmetric generating set $T \subset H$ and some $C>0$ such that $$ d_{T}^{\prime}(y,h)-C \le d_{S}(y,h) \le d_{T}^{\prime}(y,h)+C \;\; \forall y,h\in H $$ is satisfied? (Edit: Solved as negative by Michael Stoll - Take $G=\mathbb{Z}$ with $S=\{\pm 1\}$ and $H=2\mathbb{Z}$).

  2. Can one find some symmetric generating sets $ S \subset G $ and $ T \subset H $ and some $C>0$ such that $$ d_{T}^{\prime}(y,h)-C \le d_{S}(y,h) \le d_{T}^{\prime}(y,h)+C \;\; \forall y,h\in H $$ is satisfied?

  3. Does the answer to one of the prior questions change if $H$ is assumed to be nilpotent? (Edit: part 1 is solved by Michael Stoll with the same example above)

Edit: As a response to Michael's example, I would like to raise another question:

  1. Does the answer to one of the prior questions change if we relax our demand, replacing the wanted inequality by $$ Kd_{T}^{\prime}(y,h)-C \le d_{S}(y,h) \le Kd_{T}^{\prime}(y,h)+C \;\; \forall y,h\in H $$ for some constants $C,K>0$?
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    $\begingroup$ The answer to Question 1 is "No". Take $G = \mathbb Z$ and $S = \{\pm 1\}$, $H = 2{\mathbb Z}$. For any $h \in H$, $|h|_S = |h|$, but $|h|_T$ will grow like $|h|/\max\{|t| : t \in T\} \le |h|/2$. $\endgroup$ – Michael Stoll Nov 16 '14 at 15:42
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    $\begingroup$ Here is an example of a positive answer to Question 4. Take $G = \langle x, y \rangle$, the free group on two generators, with standard generating set $S = \{x,x^{-1},y,y^{-1}\}$. Take for $H$ the "even" subgroup (whose elements are represented by words of even length). Then with $T$ the set of all elements of length 2, the inequality holds with $K = 2$, $C = 0$ (i.e., it is actually an equality). The same construction will work when all relations among elements of $S$ have even length and $H$ is the subgroup of elements of even length. $\endgroup$ – Michael Stoll Nov 16 '14 at 16:50
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    $\begingroup$ Here is another situation in which even 2 is true as an equality. Suppose that $H$ is a normal subgroup and that there is a finite subgroup $K$ such that $G = HK$. Let $T$ be any symmetric generating set of $H$ that is closed under conjugation by elements from $K$ and contains $H \cap K \setminus \{1\}$. Let $S = T \cup K \setminus \{1\}$. Since $T \subset S$, we trivially have $|h|_T \ge |h|_S$. On the other hand, any $S$-word $t_0 k_1 t_1 k_2 \cdots k_n t_n$ (with $T$-words $t_j$ and $k_j \in K$) can be rewritten as an $S$-word $t'_0 t'_1 \cdots t'_n k_1 \cdots k_n$ of the same length. ... $\endgroup$ – Michael Stoll Nov 18 '14 at 19:10
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    $\begingroup$ ... If the result is in $H$, then $k_1 k_2 \cdots k_n =: k \in H$, so the element can be written as a $T$-word of at most the same length (either $k = 1$, then the $K$-part can be left out, or else $n$ must be $\ge 1$, and the $K$-part can be replaced by one generator $k \in K \cap H \setminus \{1\} \subset T$). This shows $|h|_T \le |h|_S$ for $h \in H$. $\endgroup$ – Michael Stoll Nov 18 '14 at 19:11
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    $\begingroup$ Re the index 2 case: fix any symmetric generating set $T_0$ of $H$ such that $1 \in T_0$ and an element $g \in G \setminus H$. Take $$S = \{gt : t \in T_0\} \cup \{tg^{-1} : t \in T_0\}.$$ Then $S$ is a symmetric generating set of $G$ (we can get any $t \in T_0$ from a length-2 word in $S$, so we can generate $H$, and we get the coset $gH$) and an $S$-word represents an element of $H$ if and only if its length is even. $\endgroup$ – Michael Stoll Nov 19 '14 at 20:00
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(Edit:) This answers a different question, without the "finite index" assumption.

I think I can answer all three questions in the negative. Take for $G$ the Heisenberg group $$ G = \left(\begin{matrix} 1 & {\mathbb Z} & {\mathbb Z} \\ 0 & 1 & {\mathbb Z} \\ 0 & 0 & 1\end{matrix}\right) $$ and for $H$ its center (which is the $\mathbb Z$ sitting in the upper right corner). Then for any symmetric generating set $S$ of $G$ and any $z \in {\mathbb Z} \cong H$, we have $|z|_S \ll \sqrt{|z|}$, whereas for any symmetric generating set $T$ of $H$, we have $|z|_T \gg |z|$. Note that both $G$ and $H$ are nilpotent.

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  • $\begingroup$ Thanks, but I'm interested in the case when $H$ is of finite index inside $G$, as written in the beginning. $\endgroup$ – Miel Sharf Nov 16 '14 at 15:52
  • $\begingroup$ OK, sorry, I had overlooked this assumption. The counter-example to 1 is still vaild, though. $\endgroup$ – Michael Stoll Nov 16 '14 at 15:54
  • $\begingroup$ Could you please elaborate more? $\endgroup$ – Miel Sharf Nov 16 '14 at 15:56
  • $\begingroup$ The counter-example is in the comment to yor question. Can you be more precise as to what you would like to see more details on? $\endgroup$ – Michael Stoll Nov 16 '14 at 15:59

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