# 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.

shehu mundi

August 18@ 10:49 amgiven a set{(0,1),(1,0),(1,1),(1,2),(1,3),(3,2),(3,0)…}show that the above pairs of integers modulo n is a cyclic group