Question

Prove that a k-cycle in the group S_n has order k.

Short answer

It is proved that a k-cycle in the group S_n has order k.

Step-by-step solution

Definition of cyclic group

A cyclic group can be generated by a single element. That single element is known as a generator of the group.

Proving that a cycle in a group  has order k

Suppose there is an arbitrary element c$\in S_n$

For any$1\le i\le k$ and k cycle

We can write transposition as,

$\left({\mathrm{a}}_1....{\mathrm{a}}_{\mathrm{k}}\right)\mathrm{c}=\left\{\begin{array}{cc}{\mathrm{a}}_{\mathrm{i}+1}& \mathrm{i}\in \left[1....\mathrm{k}-1\right]:\mathrm{c}={\mathrm{a}}_{\mathrm{i}}\\ {\mathrm{a}}_{\mathrm{i}}& \mathrm{c}={\mathrm{a}}_{\mathrm{k}}\\ \mathrm{c}& \mathrm{i}\in \left[1.......\mathrm{k}\right]:\mathrm{c}\ne {\mathrm{a}}_{\mathrm{i}}\end{array}\right\}$

Now, if we consider the identity element,$1\le j\le k$

${\left({\mathrm{a}}_1.......{\mathrm{a}}_{\mathrm{k}}\right)}^{\mathrm{j}}\mathrm{c}=\left\{\begin{array}{cc}{\mathrm{a}}_{\mathrm{i}+\mathrm{j}},& \mathrm{i}\in \left[1....\mathrm{k}\right]:\mathrm{i}+\mathrm{j}\le \mathrm{k}\mathrm{and}\mathrm{c}={\mathrm{a}}_{\mathrm{i}}\\ {\mathrm{a}}_{\mathrm{i}+\mathrm{j}-\mathrm{k}},& \mathrm{i}\in \left[1...\mathrm{k}\right]:\mathrm{i}+\mathrm{j}>\mathrm{k}\mathrm{and}\mathrm{c}={\mathrm{a}}_{\mathrm{i}}\\ \mathrm{c},& \mathrm{i}\in \left[1....\mathrm{k}\right]:\mathrm{c}\ne {\mathrm{a}}_{\mathrm{i}}\end{array}\right\}$

From the above expression, it can be seen that for j=k, data-custom-editor="chemistry" ${\left({\mathrm{a}}_1.....{\mathrm{a}}_{\mathrm{k}}\right)}^j$is an identity. Thus, from the definition of a cyclic function, we can conclude that a k cycle in the group S_n has order k.