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 the 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 $n$ and only $n$ distinct elements in the cyclic group, i.e. the order of the group is $n$.

Theorem 3: The generators 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.