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I started reading with growth of groups. The problem is the following:

Problem : Let $G$ be an infinite finitely generated group. a. If $G$ is polynomial growth then so is $G^m$ (a direct product of $G$). Moreover, the growth function $\gamma_G$ is not equivalent of $\gamma_{G^m}$ unless $m=1$. b. If $G$ is exponential growth then so is $G^m$, and their growth functions are equivalent.

Could any one give me a hint. Thanks in advance.

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Maybe try showing that $\gamma_{G^m}=\gamma_G^m$. – user641 Nov 20 '12 at 10:14
up vote 1 down vote accepted

Let $S= \{e_1, \dots ,e_r \}$ be a generator set of $G$. For $G^m$, take the generator set $S_m= \{(e_1,1),\dots,(e_r,1),(1,e_1) ,\dots ,(1,e_r)\}$.

Claim: For $m=2$, $\ell g(x,y)=\ell g(x)+ \ell g(y)$.

If $x=w_1$ and $y=w_2$ where $w_1$ and $w_2$ are words over $S$, then $(x,y)=(w_1,1) \cdot (1,w_2)$, so $\ell g(x,y) \leq \ell g(x)+ \ell g(y)$.

If $(x,y)=w_3$ where $w_3$ is a word over $S_2$, you can write $w_3=(w_1,1) \cdot (1,w_2)$ where $w_1$ and $w_2$ are words over $S$ (since $G \times \{1\}$ and $\{1\} \times G$ commute and have trivial intersection); in particular, $\ell g(w_3)=\ell g(w_1)+ \ell g(w_2)$. Therefore, $\ell g(x,y) \geq \ell g(x)+ \ell g(y)$.

Now by induction, you can show that:

Lemma: $\displaystyle \ell g(x_1, \dots ,x_m)= \sum\limits_{i=1}^m \ell g(x_i)$.

You deduce that $\left\{ \begin{array}{ccc} B_{G^m}(k) & \to & B_{G}(k)^m \\ (x_1,\dots,x_m) & \mapsto & (x_1,\dots,x_m) \end{array} \right.$ and $\left\{ \begin{array}{ccc} B_{G}(k)^m & \to & B_{G^m}(km) \\ (x_1,\dots,x_m) & \mapsto & (x_1,\dots,x_m) \end{array} \right.$ are well-defined and injective, hence: $$\gamma_{G^m}(k/m) \leq \gamma_{G^m}(k) \leq \gamma_G(k)^m \leq \gamma_{G^m}(km).$$

Consequently, $\gamma_{G}^m \sim \gamma_{G^m}$. Then the given assertions follow.

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