Proof of the connection of the growth functions of a residually finite group and all of its finite quotients

Question

I was reading the research article entitled "Asymptotic growth of finite groups" by Sarah Black. Professor Black makes the following statement at the bottom of page 406:

Indeed, given a f.g. residually finite group $\Gamma$ and finite generating set $S$, consider the family $\mathcal{G} = (G_i)$, consisting of all finite quotients of $\Gamma$, with respective generating sets $\mathcal{S}$, where, for each i, $S _i$ is the epimorphic image of $S$ under the mapping $\Gamma \twoheadrightarrow G_i$. Then it is easy to see that $\gamma_S^{\Gamma}$ and $\gamma_{\mathcal{S}}^{\mathcal{G}}$ represent the same function.

Notation. Following the notation of [1, Section 1], for $n \in \mathbb{N}$ we have $$\gamma_S^{\Gamma}(n) := \left\vert \{g \in \Gamma \,\right\vert\, l_S(g) \le n \}\vert, \,\, \gamma_{\mathcal{S}}^{\mathcal{G}}(n) := \max_i \gamma_{S_i}^{G_i}(n).$$

If someone helps me get a detailed proof of the above statement, then I will be grateful to him/her forever.

---

[1] S. Black, "Asymptotic growth of finite groups", 1997.

Answer 1

Sarah Black's statement is an immediate consequence of the following claim:

Claim. Let $n \in \mathbb{N}$. Keeping Black's notation in mind, let us set $B^{\Gamma}_S(n) = \{g \in \Gamma \,\vert\, l_S(g) \le n\}$. Then there is an index $i = i(n)$ such that the epimorphism $\Gamma \twoheadrightarrow G_i$ mapping element-wise $S$ to $S_i$ restricts to an isometry from $B^{\Gamma}_S(n)$ onto $B^{G_i}_{S_i}(n)$.

Proof. Since $\Gamma$ is residually finite, we can find for every $g \in \Gamma \setminus \{1\}$ an index $i(g)$ such that $g$ maps to a non-trival element under the mapping $\pi_g: \Gamma \twoheadrightarrow G_{i(g)}$ which is defined by $\pi_g(S) = S_{i(g)}$. Let $n > 0$ (the result is obvious if $n = 0$) and consider the image of $\Gamma$ in the direct product $\Pi_{g \in B_S^{\Gamma}(2n) \setminus \{1\}} G_{i(g)}$ under the product map $\pi := \times_{g \in B_S^{\Gamma}(2n) \setminus \{1\}} \pi_g$. This image is a finite quotient $G_{i(n)}$ of $\Gamma$ and the restriction of $\pi$ to $B_S^{\Gamma}(n)$ is injective by construction.