Let $\mathbb{Z}^k$ denote the finitely-generated free abelian group on $k$ generators and consider an injective homomorphism $$\chi \colon (\mathbb{Z}^k,+) \to (\mathbb{R},+).$$ Now let $(a_n)$ be a sequence of pairwise distinct elements in $\mathbb{Z}^k$. My question is:
Does there exist a subsequence of $(a_n)$ such that $\vert \chi(a_n) \vert$ tends to $+\infty$?
The answer for $k=1$ is yes as then $\chi$ corresponds to a non-zero multiplication and one can extract a subsequence of $(a_n) \subseteq \mathbb{Z}$ either going towards $+ \infty$ of $-\infty$.
For $k>1$ this seems more subtle and I'm not sure anymore whether the answer to the question is affermative. Maybe the following bit might be helpful:
Observe that $\chi$ can be viewed as a vector $(\chi_1, \dots, \chi_k) \in \mathbb{R}^k$ such that $$\chi(b^1,\dots b^k)=\langle (\chi_1, \dots,\chi_k),(b^1,\dots,b^k) \rangle=\sum_{i=1}^k \chi_i \cdot b^i.$$ Injectivity of $\chi$ then implies that the $(\chi_i)_{i=1}^k \subseteq \mathbb{R}$ are linearly independent over $\mathbb{Q}$.
I originally just left a comment since I only had a general concept and didn't have the time to work out the details. But since comments are ephemeral, and I've since put more thought into it, I am giving a proper answer.
Let $$\eta = \sum_{k=0}^\infty 2^{-2^k}$$
Since its binary expansion does not repeat, $\eta$ must be irrational. for all $n\ge 0$, let $$q_n = 2^{2^n},\quad p_n = \lfloor q_n\eta \rfloor$$ Then $$0 \le q_n\eta - p_n = 2^{2^n}\sum_{k=n+1}^\infty 2^{-2^k} < 2^{2^n}\sum_{l=2^{n+1}}^\infty 2^{-l} = 2^{2^n}2^{1-2^{n+1}} = 2^{1-2^n} \to 0$$ as $n \to \infty$.
Therefore the map $\chi : \Bbb Z^2 \to \Bbb R : (n, m) \mapsto n + m\eta$ is an injective homorphism of additive groups, and there exists a sequence $(a_n) = (-p_n) \times (q_n)$ such that $\chi(a_n) \to 0$ as $n \to \infty$, and therefore the same is true of all subsequences. The pairwise distinct criterion follows from $q_n = 2^{2^n}$.