Another useful result describes the cycle structure of a power of a cycle:
.
Lemma
For each natural number 
the power 
consists
of exactly
disjoint cyclic factors, they all are
of length 
Proof: Let 
denote the length of the cyclic
factor of containing 
. This cycle is
where 
denotes the residue class of 
modulo 
.
Correspondingly the cyclic factor containing 
(if 
) must be
, so that all the cyclic factors of
have the same length . Thus is
the order of , an element of the group
which is of order . Hence divides
and 
divides 
and
therefore also .
But
so that also must divide , which proves that
in fact 
.
An easy application of 
gives
.
Corollary
The elements of order 
in the group generated
by the cycle 
are the powers 
where 
is
of the form 
and 
is relatively prime to 
A direct consequence of this is
These 
.
Corollary
The group 
contains, for each divisor of exactly one subgroup 
of
order . Furthermore, this subgroup contains 
elements
consisting of -cycles only,
if 
denotes
the Euler function 
elements form the set of generators of 
As finite cyclic groups of the same order are isomorphic, they have the same properties:
.
Corollary
A finite cyclic group 
has,
for each divisor of its order 
exactly one subgroup of
order . Furthermore, this subgroup contains generators, and
so, has exactly elements of this particular order. Moreover
Exercises
E 
.
Show that, for each 
and 
, the permutations
and 
are conjugate, if and
only if
and each length of a cyclic factor of are relatively prime.
E 
.
Prove that the invertibility of the matrix
is equivalent to the following fact:
Two elements 
are equivalent if and only if,
for each 
the number of cyclic factors of and of 
are equal:
(Later on we shall return to this and give a proof of the regularity of
We shall in fact show that the determinant of this
matrix is 
)