Question

Show that $U_{26}/〈5〉$is isomorphic to ${\mathbb{Z}}_3$ .

Short answer

The group $U_{26}/〈5〉\cong {\mathbb{Z}}_3$.

Step-by-step solution

Normal subgroup and Quotient group

Let $N$ be the normal subgroup of $G$. Then:

1. $G/N$is a group under the operation defined by $\left(Na\right)\left(Nc\right)=Nac$.

2. If $G$is finite, then the order of $G/N$is $\left|G\right|/\left|N\right|$.

3. If $G$is an abelian group, then so is $G/N$.

The group $G/N$ is called the quotient group or factor group of $G$ by $N$.

Quotient group  U26/5

The elements in the group $U_{26}$ are$\left\{1,3,5,7,9,11,15,17,19,21,23,25\right\}$, and the elements in the cyclic subgroup $〈5〉$ are$\left\{1,5,21,25\right\}$ .

Here, the order of group $U_{26}$ is 12, and the subgroup $〈5〉$ is 4, such that the order of quotient subgroup $U_{26}/〈5〉$ is 3.

U26/5≅ℤ3

Recall the theorem 8.7. Let p be a positive prime integer. Every order of group p is isomorphic to ${\mathbb{Z}}_p$.

Here, the order of $U_{26}/〈5〉$ is a positive prime integer 3, then by the theorem,$U_{26}/〈5〉\cong {\mathbb{Z}}_3$.

Therefore, the group$U_{26}/〈5〉\cong {\mathbb{Z}}_3$.