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Lemma·L33

Elements of a finite group have a finite order

In a finite group some positive power of every element is the identity.

Let be a group with finite, and . Then some positive power of is the identity: with powers from D043 and from D029.
In words
If the group has only finitely many elements and g is any of its elements, then there is an exponent n: n is a natural number, and n is positive, and multiplying g by itself n times returns to the identity.
Never needed: F10 · A03 · A04 · A05 · A07 (computed from the citation graph, not asserted).

Proof

  1. 1
    Let (D035) and fix a bijection (D034). Let be the restriction of (D043) to , so for each ; these are the powers .
  2. 2
    Suppose were injective. Then is injective (L02 (ii)). But (L15 (vi)), so T12 rules out any injection . Hence is not injective: there are with and .
  3. 3
    By L16 we may take . Then for some with (L17 (ii)), so .
  4. 4
    Now by L32 (i). Since also (D038 (G2)), left cancellation L28 (iii) applied to gives . With , this is the required positive power.

Remarks

The heart is pigeonhole: among the powers two must coincide in a group of elements, and dividing them out yields a power equal to . This is exactly what makes the order of an element well defined. In an infinite group the conclusion can fail: the integers under addition, once constructed, have elements of no finite order.