Question

Let R be a finite ring. Show that there exists an infinite sequence n1 < n2 < n3 < • • • of natural numbers such that, for any x ∈ R, we have x^n1 = x^n2 = x^n3 = • • • .

Answer 1

Label the elements of $R$ as $x_{1}, x_{2}, \ldots, x_{k}$. Since there are at most $k$ distinct elements in the set $\{x_{1}, x_{1}^{2}, \ldots\}$, there must exist

$$\begin{array}{l} r_{1} < r_{2} < \cdots \text{ such that } \\ x_{1}^{r1} = x_{1}^{r2} = \cdots. \end{array}$$

By considering $\{x_{2}^{r1}, x_{2}^{r2}, \ldots\}$, we see similarly that there exist a subsequence $s_1 < s_2 < \cdots$ of $\{r_i\}$ such that $x_{2}^{s1} = x_{2}^{s2} = \cdots$.

Repeating this construction a finite number of times, we arrive at a sequence

$$n_{1} < n_{2} < \dots \text{ such that }$$ $$x_{i}^{n1} = x_{i}^{n2} = \cdots \quad \text{ for } 1 \leq i \leq k.$$