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 (i₁ ...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 (i₁ ...i_r )^m containing i₁. This cycle is

(i₁ 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 i₂ (if r>1) must be (i₂ ...i _{bar ((l-1)m+2)}), so that all the cyclic factors of (i₁...i_r)^m have the same length l. Thus l is the order of (i₁ ...i_r)^m, an element of the group á(i₁ ...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

(i₁ ...i_r)^{m ·(r/ gcd (r,m))}=(i₁ ...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 Îd | 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.

Subgroups of Cyclic Groups