91Ó°ÊÓ

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$A^3=1=e^{i2\mathrm{Ï€}}$obtain the solution as $A=e^{i2\mathrm{Ï€}/3}$.

Therefore, the elements of this group are

$e^{i2\mathrm{Ï€}/3},e^{i4\mathrm{Ï€}/3},e^{i2\mathrm{Ï€}}=1$

See that a unit element. Their multiplication gives

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

This means that closure, and the inverse element, that is,$e^{i2\mathrm{Ï€}/3}$ is the inverse of $e^{i4\mathrm{Ï€}/3}$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 ${\mathrm{A}}^6=1={\mathrm{e}}^{\mathrm{i}2\mathrm{Ï€}}$obtain the solution as$A=e^{i\mathrm{Ï€}/3}$. Therefore, the elements of this group are

${\mathrm{e}}^{\mathrm{¾\pm Ï€}/3},{\mathrm{e}}^{\mathrm{i}2\mathrm{Ï€}/3},{\mathrm{e}}^{\mathrm{¾\pm Ï€}},{\mathrm{e}}^{\mathrm{i}4\mathrm{Ï€}/3},{\mathrm{e}}^{\mathrm{i}5\mathrm{Ï€}/3},{\mathrm{e}}^{\mathrm{i}2\mathrm{Ï€}}=1$

See that a unit element.

The multiplication of various elements gives

$$\begin{gathered} {\mathrm{e}}^{\mathrm{¾\pm Ï€}/3},{\mathrm{e}}^{\mathrm{¾\pm Ï€}/3}={\mathrm{e}}^{i2\mathrm{Ï€}/3},{\mathrm{e}}^{\mathrm{¾\pm Ï€}/3},{\mathrm{e}}^{\mathrm{i}2\mathrm{Ï€}/3} \\ ={\mathrm{e}}^{\mathrm{¾\pm Ï€}},{\mathrm{e}}^{\mathrm{¾\pm Ï€}/3},{\mathrm{e}}^{\mathrm{¾\pm Ï€}} \\ ={\mathrm{e}}^{\mathrm{i}4\mathrm{Ï€}/3}{\mathrm{e}}^{\mathrm{¾\pm Ï€}/3}{\mathrm{e}}^{\mathrm{i}4\mathrm{Ï€}/3} \\ ={\mathrm{e}}^{\mathrm{i}5\mathrm{Ï€}/3}{\mathrm{e}}^{\mathrm{¾\pm Ï€}/3}{\mathrm{e}}^{\mathrm{i}5\mathrm{Ï€}/3} \end{gathered}$$

Solve further

$$\begin{gathered} {\mathrm{e}}^{i2\mathrm{Ï€}}=1 \\ {\mathrm{e}}^{\mathrm{i}2\mathrm{Ï€}/3}{\mathrm{e}}^{\mathrm{i}2\mathrm{Ï€}/3}={\mathrm{e}}^{\mathrm{i}4\mathrm{Ï€}/3} \\ {\mathrm{e}}^{\mathrm{i}2\mathrm{Ï€}/3}{\mathrm{e}}^{\mathrm{¾\pm Ï€}}={\mathrm{e}}^{\mathrm{i}5\mathrm{Ï€}/3} \\ {\mathrm{e}}^{\mathrm{i}2\mathrm{Ï€}/3}{\mathrm{e}}^{\mathrm{i}4\mathrm{Ï€}/3}={\mathrm{e}}^{\mathrm{i}2\mathrm{Ï€}}=1 \end{gathered}$$

Solve further

$$\begin{gathered} {\mathrm{e}}^{\mathrm{i}2\mathrm{Ï€}/3}{\mathrm{e}}^{\mathrm{i}5\mathrm{Ï€}/3}={\mathrm{e}}^{\mathrm{¾\pm Ï€}/3} \\ {\mathrm{e}}^{\mathrm{¾\pm Ï€}}{\mathrm{e}}^{\mathrm{¾\pm Ï€}}={\mathrm{e}}^{\mathrm{i}2\mathrm{Ï€}}=1 \\ {\mathrm{e}}^{\mathrm{¾\pm Ï€}}{\mathrm{e}}^{\mathrm{i}4\mathrm{Ï€}/3}={\mathrm{e}}^{\mathrm{¾\pm Ï€}/3} \\ {\mathrm{e}}^{\mathrm{¾\pm Ï€}}{\mathrm{e}}^{\mathrm{i}5\mathrm{Ï€}/3}={\mathrm{e}}^{\mathrm{i}2\mathrm{Ï€}/3} \\ {\mathrm{e}}^{\mathrm{i}4\mathrm{Ï€}/3}{\mathrm{e}}^{\mathrm{i}5\mathrm{Ï€}/3}={\mathrm{e}}^{\mathrm{¾\pm Ï€}} \end{gathered}$$

${\mathrm{e}}^{\mathrm{i}2\mathrm{Ï€}/3}$is the inverse of ${\mathrm{e}}^{\mathrm{i}4\mathrm{Ï€}/3}$and vice versa, ${\mathrm{e}}^{\mathrm{¾\pm Ï€}/3}$is the inverse of ${\mathrm{e}}^{\mathrm{i}5\mathrm{Ï€}/3}$and vice versa,I and ${\mathrm{e}}^{\mathrm{¾\pm Ï€}}$, are their own inverse. Since these elements are numbers, they are associative.

Write the table of multiplication