Chapter 6.3, Problem 8E

(a)

To determine

To find: the order of element 3 in the group $\left({\mathbb{Z}}_4,+\right)$ .

Answer to Problem 8E

The order of the element 3 of the group $\left({\mathbb{Z}}_4,+\right)$ is 4.

Explanation of Solution

Concept used:

The order of an element ‘a’ of an additive group is defined as the least positive integer n such that $na=0$ , where 0 is the identity element in the additive group.

Calculation:

Consider the group $\left({\mathbb{Z}}_4,+\right)$ .

  ${\mathbb{Z}}_4=\left\{0,1,2,3,4\right\}$

Now, for the element 3 of the group $\left({\mathbb{Z}}_4,+\right)$ .

  $\begin{array}{l}1\left(3\right)=3=3\left(\mathrm{mod}4\right)\\ 2\left(3\right)=6=2\left(\mathrm{mod}4\right)\\ 3\left(3\right)=9=1\left(\mathrm{mod}4\right)\\ 4\left(3\right)=12=0\left(\mathrm{mod}4\right)\end{array}$

Thus, the order of the element 3 of the group $\left({\mathbb{Z}}_4,+\right)$ is 4.

(b)

To determine

To find: the order of element 3 in the group $\left({\mathbb{Z}}_5,+\right)$ .

Answer to Problem 8E

The order of the element 3 of the group $\left({\mathbb{Z}}_5,+\right)$ is 5.

Explanation of Solution

Concept used:

The order of an element ‘a’ of an additive group is defined as the least positive integer n such that $na=0$ , where 0 is the identity element in the additive group.

Calculation:

Consider the group $\left({\mathbb{Z}}_5,+\right)$ .

  ${\mathbb{Z}}_5=\left\{0,1,2,3,4,5\right\}$

Now, for the element 3 of the group $\left({\mathbb{Z}}_5,+\right)$ .

  $\begin{array}{l}1\left(3\right)=3=3\left(\mathrm{mod}5\right)\\ 2\left(3\right)=6=1\left(\mathrm{mod}5\right)\\ 3\left(3\right)=9=4\left(\mathrm{mod}5\right)\\ 4\left(3\right)=12=2\left(\mathrm{mod}5\right)\\ 5\left(3\right)=15=0\left(\mathrm{mod}5\right)\end{array}$

Thus, the order of the element 3 of the group $\left({\mathbb{Z}}_5,+\right)$ is 5.

(c)

To determine

To find: the order of element 3 in the group $\left({\mathbb{Z}}_6,+\right)$ .

Answer to Problem 8E

The order of the element 3 of the group $\left({\mathbb{Z}}_6,+\right)$ is 2.

Explanation of Solution

Concept used:

The order of an element ‘a’ of an additive group is defined as the least positive integer n such that $na=0$ , where 0 is the identity element in the additive group.

Calculation:

Consider the group $\left({\mathbb{Z}}_6,+\right)$ .

  ${\mathbb{Z}}_6=\left\{0,1,2,3,4,5,6\right\}$

Now, for the element 3 of the group $\left({\mathbb{Z}}_6,+\right)$ .

  $\begin{array}{l}1\left(3\right)=3=3\left(\mathrm{mod}6\right)\\ 2\left(3\right)=6=0\left(\mathrm{mod}6\right)\\ 3\left(3\right)=9=3\left(\mathrm{mod}6\right)\\ 4\left(3\right)=12=0\left(\mathrm{mod}6\right)\\ 5\left(3\right)=15=3\left(\mathrm{mod}6\right)\\ 6\left(3\right)=18=0\left(\mathrm{mod}6\right)\end{array}$

Thus, the order of the element 3 of the group $\left({\mathbb{Z}}_6,+\right)$ is 2.

(d)

To determine

To find: the order of element 3 in the group $\left({\mathbb{Z}}_8,+\right)$ .

Answer to Problem 8E

The order of the element 3 of the group $\left({\mathbb{Z}}_8,+\right)$ is 8.

Explanation of Solution

Concept used:

The order of an element ‘a’ of an additive group is defined as the least positive integer n such that $na=0$ , where 0 is the identity element in the additive group.

Calculation:

Consider the group $\left({\mathbb{Z}}_8,+\right)$ .

  ${\mathbb{Z}}_8=\left\{0,1,2,3,4,5,6,7,8\right\}$

Now, for the element 3 of the group $\left({\mathbb{Z}}_8,+\right)$ .

  $\begin{array}{l}1\left(3\right)=3=3\left(\mathrm{mod}8\right)\\ 2\left(3\right)=6=6\left(\mathrm{mod}8\right)\\ 3\left(3\right)=9=1\left(\mathrm{mod}8\right)\\ 4\left(3\right)=12=4\left(\mathrm{mod}8\right)\\ 5\left(3\right)=15=7\left(\mathrm{mod}8\right)\\ 6\left(3\right)=18=2\left(\mathrm{mod}8\right)\\ 7\left(3\right)=21=5\left(\mathrm{mod}8\right)\\ 8\left(3\right)=24=0\left(\mathrm{mod}8\right)\end{array}$

Thus, the order of the element 3 of the group $\left({\mathbb{Z}}_8,+\right)$ is 8.

(e)

To determine

To find: the order of element 3 in the group $\left({\mathbb{Z}}_9,+\right)$ .

Answer to Problem 8E

The order of the element 3 of the group $\left({\mathbb{Z}}_9,+\right)$ is 3.

Explanation of Solution

Concept used:

The order of an element ‘a’ of an additive group is defined as the least positive integer n such that $na=0$ , where 0 is the identity element in the additive group.

Calculation:

Consider the group $\left({\mathbb{Z}}_9,+\right)$ .

  ${\mathbb{Z}}_9=\left\{0,1,2,3,4,5,6,7,8,9\right\}$

Now, for the element 3 of the group $\left({\mathbb{Z}}_9,+\right)$ .

  $\begin{array}{l}1\left(3\right)=3=3\left(\mathrm{mod}9\right)\\ 2\left(3\right)=6=6\left(\mathrm{mod}9\right)\\ 3\left(3\right)=9=0\left(\mathrm{mod}9\right)\\ 4\left(3\right)=12=3\left(\mathrm{mod}9\right)\\ 5\left(3\right)=15=6\left(\mathrm{mod}9\right)\\ 6\left(3\right)=18=0\left(\mathrm{mod}9\right)\\ 7\left(3\right)=21=3\left(\mathrm{mod}9\right)\\ 8\left(3\right)=24=6\left(\mathrm{mod}9\right)\\ 9\left(3\right)=27=0\left(\mathrm{mod}9\right)\end{array}$

Thus, the order of the element 3 of the group $\left({\mathbb{Z}}_9,+\right)$ is 3.

(f)

To determine

To find: the order of element 3 in the group $\left({\cup }_5,·\right)$ .

Answer to Problem 8E

The order of the element 3 of the group $\left({\cup }_5,·\right)$ is 4.

Explanation of Solution

Concept used:

The order of an element ‘a’ of a multiplicative group is defined as the least positive integer n such that $a^n=1$ , where 1 is the identity element in the multiplicative group.

Calculation:

Consider the group $\left({\cup }_5,·\right)$ .

  ${\cup }_5=\left\{1,2,3,4\right\}$

Now, for the element 3 of the group $\left({\cup }_5,·\right)$ .

  $\begin{array}{l}{3}^1=3=3\left(\mathrm{mod}5\right)\\ 3^2=9=4\left(\mathrm{mod}5\right)\\ 3^3=27=2\left(\mathrm{mod}5\right)\\ 3^4=81=1\left(\mathrm{mod}5\right)\end{array}$

Thus, the order of the element 3 of the group $\left({\cup }_5,·\right)$ is 4.

(g)

To determine

To find: the order of element 3 in the group $\left({\cup }_7,·\right)$ .

Answer to Problem 8E

The order of the element 3 of the group $\left({\cup }_7,·\right)$ is 6.

Explanation of Solution

Concept used:

The order of an element ‘a’ of a multiplicative group is defined as the least positive integer n such that $a^n=1$ , where 1 is the identity element in the multiplicative group.

Calculation:

Consider the group $\left({\cup }_7,·\right)$ .

  ${\cup }_7=\left\{1,2,3,4,5,6\right\}$

Now, for the element 3 of the group $\left({\cup }_7,·\right)$ .

  $\begin{array}{l}{3}^1=3=3\left(\mathrm{mod}7\right)\\ 3^2=9=2\left(\mathrm{mod}7\right)\\ 3^3=27=6\left(\mathrm{mod}7\right)\\ 3^4=81=4\left(\mathrm{mod}7\right)\\ 3^5=243=5\left(\mathrm{mod}7\right)\\ 3^6=729=1\left(\mathrm{mod}7\right)\end{array}$

Thus, the order of the element 3 of the group $\left({\cup }_7,·\right)$ is 6.

(h)

To determine

To find: the order of element 3 in the group $\left({\cup }_{11},·\right)$ .

Answer to Problem 8E

The order of the element 3 of the group $\left({\cup }_{11},·\right)$ is 5.

Explanation of Solution

Concept used:

The order of an element ‘a’ of a multiplicative group is defined as the least positive integer n such that $a^n=1$ , where 1 is the identity element in the multiplicative group.

Calculation:

Consider the group $\left({\cup }_{11},·\right)$ .

  ${\cup }_{11}=\left\{1,2,3,4,5,6,7,8,9,10\right\}$

Now, for the element 3 of the group $\left({\cup }_{11},·\right)$ .

  $\begin{array}{l}{3}^1=3=3\left(\mathrm{mod}11\right)\\ 3^2=9=9\left(\mathrm{mod}11\right)\\ 3^3=27=5\left(\mathrm{mod}11\right)\\ 3^4=81=4\left(\mathrm{mod}11\right)\\ 3^5=243=1\left(\mathrm{mod}11\right)\\ 3^6=729=3\left(\mathrm{mod}11\right)\\ 3^7=2187=9\left(\mathrm{mod}11\right)\end{array}$

Thus, the order of the element 3 of the group $\left({\cup }_{11},·\right)$ is 5.