Question

9. Show that$U_{15}$ is generated by the set $\left\{2,13\right\}$.

Short answer

The group$U_{15}$ can be generated by the set $\left\{2,13\right\}$.

Step-by-step solution

Cyclic Subgroup

Every element x in a group G, form a subgroup of all integer powers is called a cyclic subgroup of x.

Elements in U15

The set of elements in $U_{15}$is $U_{15}=\left\{1,2,4,7,8,11,13,14\right\}$.

Generating the group U15 by the set 2,13

For element $2\in U_{15},2^1=2,2^2=4,2^3=8,2^4=16=1$. Thus, the cyclic subgroup generated by element 2 is .

For element $13\in U_{15},{13}^1=13,{13}^2=169=4,{13}^3=7,{13}^4=1$. Thus, the cyclic subgroup generated by element 13 is .

Thus, the elements generated by the set$\left\{2,13\right\}$are$\left\{1,2,4,7,8,11,13,14\right\}$which is the group $U_{15}$.

Therefore, the group$U_{15}$can be generated by the set $\left\{2,13\right\}$.