| | | **Subgroups of Cyclic Groups** |

### Subgroups of Cyclic Groups

Another useful result describes the cycle structure of a power of a cycle:

**Lemma: **
*
For each natural number **m*
the power *(i*_{1} ...i_{r})^{m} consists
of exactly
* gcd (r,m)* disjoint cyclic factors, they all are
of length *r/ gcd (r,m).*

Proof: Let *l* denote the length of the cyclic
factor of *(i*_{1} ...i_{r}
)^{m} containing *i*_{1}. This cycle is

*(i*_{1} i_{ bar (m+1)} i_{ bar (2m+1)}
...i_{ bar ((l-1)m+1)}),

where * bar (k)* denotes the residue class of *k* modulo *r*.
Correspondingly the cyclic factor containing *i*_{2} (if *r>1*) must be
*(i*_{2} ...i_{ bar ((l-1)m+2)}), so that all the cyclic factors of
*(i*_{1}...i_{r})^{m} have the same length *l*. Thus *l* is
the order of *(i*_{1} ...i_{r})^{m}, an element of the group
* á(i*_{1} ...i_{r}) ñ which is of order *r*. Hence *r* divides
*l ·m* and *r/ gcd (r,m)* divides *l ·(m/ gcd (r,m))* and
therefore also *l*.
But

*(i*_{1} ...i_{r})^{m ·(r/ gcd (r,m))}=(i_{1} ...i_{r})^{r ·(m/ gcd (r,m))}=1,

so that also *l* must divide *r/ gcd (r,m)*, which proves that
in fact *l=r/ gcd (r,m)*.

An easy application of the lemma above gives

**Corollary: **
*
The elements of order **d* in the group generated
by the cycle *(1 ...n)* are the powers *(1 ...n)*^{i}, where *i* is
of the form **(**n ·j**)/(**d**)**, and *1 £j<d* is relatively prime to *d.*

A direct consequence of this is

**Corollary: **
*
The group ** C*_{m} := á(1 ...m)
ñ contains, for each divisor *d* of *m* exactly one subgroup *U* of
order *d*. Furthermore, this subgroup contains * f(d)* elements
consisting of *d*-cycles only,
if * f(-)* denotes
the *Euler function* * f(d):= | {i Î*__d__ | gcd (d,i)=1 } | .

These * f(d)* elements form the set of generators of *U.*

As finite cyclic groups of the same order are isomorphic, they have
the same properties:

**Corollary: **
*
A finite cyclic group **G* has,
for each divisor *d* of its order * | G | ,*
exactly one subgroup *U* of
order *d*. Furthermore, this subgroup contains * f(d)* generators, and
so, *G* has exactly * f(d)* elements of this particular order. Moreover
* å*_{d | n} f(d)=n.

last changed: January 19, 2005

| | | **Subgroups of Cyclic Groups** |