Lemma·L35
The cyclic subgroup generated by an element
The powers of an element form a subgroup whose size is the order of the element.
Let
be a group with
finite,
, and
(D044). The cyclic subgroup generated by
is the set of its powers
It is a subgroup of
, and
is a bijection from
onto
; hence
(D035).
In words
For an element g of a finite group with d its order, the subgroup generated by g is the set of those x in the group such that some exponent k exists, a natural number, with x equal to the k-th power of g. It is a subgroup, sending each k to the k-th power of g is a bijection from the natural number d onto this set of powers, and so its size is exactly d, the order of g.
Never needed: F10 · A03 · A04 · A05 · A07 (computed from the citation graph, not asserted).
Proof
- 1
- 2
- 3Surjective. Let send (a restriction of ). By the reduction step, every member of is some with , so is surjective.
- 4
- 5
∎
Remarks
is the smallest subgroup containing
, and it is abelian. A group that equals
for some
is cyclic. The identity
turns the order of an element into a subgroup size, so Lagrange's theorem applies to it: that is T15.