Following the text discussion of the cyclic group of order 4, and Problem 1, discuss

(a) The cyclic group of order 3 (see Chapter 2, Problem 10.32);

(b) The cyclic group of order 6.

For a cyclic group of order 3, elements $A,A^2,A^3=1.$

For a cyclic group of order 6, the elements $A,A^2,A^3,A^4,A^5,A^6=1$.

A branch of abstract algebra, a cyclic group or monogenous group is a group that is generated by a single element.

(a) In this problem discuss the cyclic group of orders 3

For a cyclic group of order 3, elements $A,A^2,A^3=1$. Since$A^3=1=e^{i2\pi }$obtain the solution as $A=e^{i2\pi /3}$.

Therefore, the elements of this group are

$e^{i2\pi /3},e^{i4\pi /3},e^{i2\pi }=1$

See that a unit element. Their multiplication gives

${\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }/3}{\mathrm{e}}^{\mathrm{i}4\mathrm{\pi }/3}={\mathrm{e}}^{\mathrm{i}6\mathrm{\pi }/3}={\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }}=1$

This means that closure, and the inverse element, that is,$e^{i2\pi /3}$ is the inverse of $e^{i4\pi /3}$and vice versa, and 1 is its own inverse. Since these elements are numbers, they are associative. Therefore, this is a group. Write the table of multiplication

(b) For a cyclic group of order 6, the elements $A,A^2,A^3,A^4,A^5,A^6=1$. Since ${\mathrm{A}}^6=1={\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }}$obtain the solution as$A=e^{i\pi /3}$. Therefore, the elements of this group are

${\mathrm{e}}^{\mathrm{i\pi }/3},{\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }/3},{\mathrm{e}}^{\mathrm{i\pi }},{\mathrm{e}}^{\mathrm{i}4\mathrm{\pi }/3},{\mathrm{e}}^{\mathrm{i}5\mathrm{\pi }/3},{\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }}=1$

See that a unit element.

The multiplication of various elements gives

$$\begin{gathered} {\mathrm{e}}^{\mathrm{i\pi }/3},{\mathrm{e}}^{\mathrm{i\pi }/3}={\mathrm{e}}^{i2\pi /3},{\mathrm{e}}^{\mathrm{i\pi }/3},{\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }/3} \\ ={\mathrm{e}}^{\mathrm{i\pi }},{\mathrm{e}}^{\mathrm{i\pi }/3},{\mathrm{e}}^{\mathrm{i\pi }} \\ ={\mathrm{e}}^{\mathrm{i}4\mathrm{\pi }/3}{\mathrm{e}}^{\mathrm{i\pi }/3}{\mathrm{e}}^{\mathrm{i}4\mathrm{\pi }/3} \\ ={\mathrm{e}}^{\mathrm{i}5\mathrm{\pi }/3}{\mathrm{e}}^{\mathrm{i\pi }/3}{\mathrm{e}}^{\mathrm{i}5\mathrm{\pi }/3} \end{gathered}$$

Solve further

$$\begin{gathered} {\mathrm{e}}^{i2\pi }=1 \\ {\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }/3}{\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }/3}={\mathrm{e}}^{\mathrm{i}4\mathrm{\pi }/3} \\ {\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }/3}{\mathrm{e}}^{\mathrm{i\pi }}={\mathrm{e}}^{\mathrm{i}5\mathrm{\pi }/3} \\ {\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }/3}{\mathrm{e}}^{\mathrm{i}4\mathrm{\pi }/3}={\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }}=1 \end{gathered}$$

Solve further

$$\begin{gathered} {\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }/3}{\mathrm{e}}^{\mathrm{i}5\mathrm{\pi }/3}={\mathrm{e}}^{\mathrm{i\pi }/3} \\ {\mathrm{e}}^{\mathrm{i\pi }}{\mathrm{e}}^{\mathrm{i\pi }}={\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }}=1 \\ {\mathrm{e}}^{\mathrm{i\pi }}{\mathrm{e}}^{\mathrm{i}4\mathrm{\pi }/3}={\mathrm{e}}^{\mathrm{i\pi }/3} \\ {\mathrm{e}}^{\mathrm{i\pi }}{\mathrm{e}}^{\mathrm{i}5\mathrm{\pi }/3}={\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }/3} \\ {\mathrm{e}}^{\mathrm{i}4\mathrm{\pi }/3}{\mathrm{e}}^{\mathrm{i}5\mathrm{\pi }/3}={\mathrm{e}}^{\mathrm{i\pi }} \end{gathered}$$

${\mathrm{e}}^{\mathrm{i}2\mathrm{\pi }/3}$is the inverse of ${\mathrm{e}}^{\mathrm{i}4\mathrm{\pi }/3}$and vice versa, ${\mathrm{e}}^{\mathrm{i\pi }/3}$is the inverse of ${\mathrm{e}}^{\mathrm{i}5\mathrm{\pi }/3}$and vice versa,I and ${\mathrm{e}}^{\mathrm{i\pi }}$, are their own inverse. Since these elements are numbers, they are associative.

Write the table of multiplication