Question

Let $\mathit{M}$ be the cyclic subgroup $\mathbf{⟨}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{⟩}$ of the additive group $\mathit{G}\mathbf{=}{\mathit{\mathbb{Z}}}_{\mathbf{2}}\mathbf{\times }{\mathit{\mathbb{Z}}}_{\mathbf{4}}$ and let $\mathit{N}$ be the cyclic subgroup $\mathbf{⟨}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{⟩}$as in example 4.Verify that $\mathit{M}$ is isomorphic to $\mathit{N}$ .

Short answer

The cyclic subgroup $M$ is isomorphic to the cyclic subgroup $N$.

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 $G/N$.

Group G  and their subgroups

Let $G={\mathbb{Z}}_2\times {\mathbb{Z}}_4$ be the group of order 8 and M be the subgroup generated by $(0,2)$, N be the subgroup generated by $(1,2)$.

The elements in M are $\left\{(0,2)(0,0)\right\}$, and the elements in N are $\left\{(1,2)(0,0)\right\}$.

 M≅N

Since the order of both the subgroups is 2, they are isomorphic.

Therefore, the cyclic subgroup $M$ is isomorphic to the cyclic subgroup $N$.