Theorem. The set of all elements of a ring, which have a finite order in the additive group of the ring, is a (two-sided) ideal of the ring.

Proof. Let $S$ be the set of the elements with finite order in the ring $R$ . Denote by $o(x)$ the order of $x$ . Take arbitrary elements $a,\,b$ of the set $S$ .

If $\lcm(o(a),\,o(b)) = n = ko(a) = lo(b)$ , then $$n(a-b) = na-nb = ko(a)a-lo(b)b = k\cdot0-l\cdot0 = 0-0 = 0.$$ Thus $o(a-b) \leqq n < \infty$ and so $a-b \in S$ .

For any element $r$ of $R$ we have $$o(a)(ra) = \underbrace{ra+ra+\ldots+ra}_{o(a)} = r(\underbrace{a+a+\ldots+a}_{o(a)}) = r(o(a)a) = r\cdot0 = 0.$$ Therefore, $o(ra) \leqq o(a) < \infty$ and $ra \in S$ . Similarly, $ar \in S$ .

Since $S$ satisfies the conditions for an ideal, the theorem has been proven.