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# ideal of elements with finite order

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

$n(a-b)=na-nb=ko(a)a-lo(b)b=k\cdot 0-l\cdot 0=0-0=0.$ |

Thus $o(a-b)\leqq n<\infty$ and so $a-b\in S$.

For any element $r$ of $R$ we have

$o(a)(ra)=\underbrace{ra+ra+\ldots+ra}_{{o(a)}}=r(\underbrace{a+a+\ldots+a}_{{o% (a)}})=r(o(a)a)=r\cdot 0=0.$ |

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.

Related:

OrderGroup, Lcm, Multiple, OrdersOfElementsInIntegralDomain, CharacteristicOfFiniteRing

Type of Math Object:

Theorem

Major Section:

Reference

Parent:

Groups audience:

## Mathematics Subject Classification

20A05*no label found*16D25

*no label found*

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