Properties of Cyclic Groups

Theorem 1: Every cyclic group is Abelian.

Proof: Let a be a generator of a cyclic group G and let {a^r},{a^s} \in G for any r,s \in I then

{a^r} \cdot {a^s} = {a^{r + s}} = {a^{s + r}} = {a^s} \cdot {a^r}

(Because r + s = s + r for r,s \in I)

Thus the operation is commutative and hence the cyclic group G is Abelian.

Note: For the addition composition the above proof could have been written as {a^r} + {a^s} = ra + sa = as + ra = {a^s} + {a^r} (addition of integer is commutative)

Thus the operation + is commutative in G.

Theorem 2: The order of a cyclic group is same as the order of its generator.

Proof: Let the order of a generator a of a cyclic group be n, then

{a^n} = e

while {a^s} \ne e for 0 < s < n

When s > n,\,\,s = nq + r,\,\,\,0 \leqslant r < n (say)

We observe that

\begin{gathered} {a^s} = {a^{nq + r}} = {\left( {{a^n}} \right)^q} \cdot {a^r} = {a^q} \cdot {a^r} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = e \cdot {a^r} = {a^r} \\ \end{gathered}

Thus there are exactly n elements in the group by {a^r}, where 0 \leqslant r < n. Therefore there are nand only ndistinct elements in the cyclic group, i.e. the order of the group is n.

Theorem 3: The generator of a cyclic group of order n are all the elements {a^p}, p being prime to n and 0 < p < n.

Proof:
We know that

{\left( {{a^p}} \right)^n} = {\left( {{a^n}} \right)^p} = e

Therefore the order of {a^p} is n.

Also {\left( {ap} \right)^s} = a{p^s} \ne e if 0 < p < n, because n does not divide p, nor does it divide s, therefore it does not divide ps.

Now let,

ps = nq + r,\,\,\,0 < r < n

 and

\begin{gathered} {\left( {{a^p}} \right)^s} = {a^{ps}} = {a^{nq + r}} \\ \,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {{a^n}} \right)^q} \cdot {a^r} = {e^q} \cdot {a^r} \\ \,\,\,\,\,\,\,\,\,\,\,\,\, = e \cdot {a^r} = {a^r} \ne 0 \\ \end{gathered}

Thus, {a^p} is a generator of the group.