Question

1. List all the cyclic subgroups

(b) ${\mathbf{U}}_{30}$

Short answer

The cyclic subgroups of ${\mathrm{U}}_{30}$are $⟨1⟩,⟨7⟩,⟨11⟩,⟨17⟩,⟨19⟩,⟨29⟩$.

Step-by-step solution

Cyclic Subgroup

Every element x in a group G, form a subgroup of all integer powers$⟨\mathrm{x}⟩=\{{\mathrm{x}}^{\mathrm{k}},\mathrm{k}\in \mathrm{\mathbb{Z}}\}$ is called a cyclic subgroup of x.

Elements in U30

The set of elements in${\mathrm{U}}_{30}$ is ${\mathrm{U}}_{30}=\{1,7,11,13,17,19,23,29\}$.

Cyclic Subgroups of U30

For element $1\in {\mathrm{U}}_{30}$, $1^1=1$and for any integer $\mathrm{k}\in \mathrm{\mathbb{Z}}$,$1^k=1$. Thus, the cyclic subgroup generated by element 1 is $⟨1⟩=\left\{1\right\}$.

For element $7\in {\mathrm{U}}_{30}$, $7^1=7,7^2=49=19,7^3=343=13,7^4=2401=1$. Thus, the cyclic subgroup generated by element 7 is $⟨7⟩=\{7,19,13,1\}$.

For element role="math" localid="1654321498298" $11\in {\mathrm{U}}_{30}$, ${11}^1=11,{11}^2=121=1$. Thus, the cyclic subgroup generated by element 11 is $⟨11⟩=\{11,1\}$.

For element $13\in {\mathrm{U}}_{30}$,${13}^1=13,{13}^2=169=19,{13}^3=7,{13}^4=1$. Thus, the cyclic subgroup generated by element 13 is $⟨13⟩=\{13,19,7,1\}=⟨7⟩$.

For element $17\in {\mathrm{U}}_{30}$, ${17}^1=17,{17}^2=19,{17}^3=23,{17}^4=1$. Thus, the cyclic subgroup generated by element 17 is $⟨17⟩=\{17,19,23,1\}$.

For element $19\in U_{30}$, ${19}^1=19,{19}^2=1$. Thus, the cyclic subgroup generated by element 19 is $⟨19⟩=\{19,1\}$.

For element $23\in {\mathrm{U}}_{30}$,${23}^1=23,{23}^2=19,{23}^3=17,{23}^4=1$. Thus, the cyclic subgroup generated by element 23 is $⟨23⟩=\{23,19,17,1\}=⟨17⟩$.

For element $29\in U_{30}$, ${29}^1=29,{29}^2=1$. Thus, the cyclic subgroup generated by element 29 is $⟨29⟩=\{29,1\}$.

Therefore, there are 6 different cyclic subgroups which are .$⟨1⟩,⟨7⟩,⟨11⟩,⟨17⟩,⟨19⟩,⟨29⟩$