Question

Is ${\mathbf{\mathbb{Z}}}_{\mathbf{8}}$isomorphic to ${\mathbf{\mathbb{Z}}}_{\mathbf{4}}\mathbf{\times }{\mathbf{\mathbb{Z}}}_{\mathbf{2}}$?

Short answer

Answer:

${\mathrm{\mathbb{Z}}}_4\times {\mathrm{\mathbb{Z}}}_2$is not isomorphic to ${\mathrm{\mathbb{Z}}}_8$.

Step-by-step solution

Cyclic Group

A group $\mathit{G}$ is called cyclic if there is an element $\mathit{g}\mathbf{\in }\mathit{G}$, which can generate the whole group that is,

$$\begin{gathered} \mathit{G}\mathbf{=}\mathbf{<}\mathbf{g}\mathbf{>} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\{}\mathbf{g}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{/}\mathbf{n}\mathbf{\in }\mathbf{Z}\mathbf{\}} \end{gathered}$$

Check ℤ8 isomorphic to ℤ4×ℤ2

Consider two groups ${\mathrm{\mathbb{Z}}}_4\times {\mathrm{\mathbb{Z}}}_2$and ${\mathrm{\mathbb{Z}}}_8$.

Find the order of each group as:

Find the order 1of ${\mathrm{\mathbb{Z}}}_2$as follows:

Order of $1$in ${\mathrm{\mathbb{Z}}}_2$is $2$.

Find the order of ${\mathrm{\mathbb{Z}}}_4$as follows:

Order of $1$and $2$in ${\mathrm{\mathbb{Z}}}_4$is $4$.

Find the order of ${\mathrm{\mathbb{Z}}}_4\times {\mathrm{\mathbb{Z}}}_2$as follows:

${\mathrm{\mathbb{Z}}}_4\times {\mathrm{\mathbb{Z}}}_2=\left\{\left(0,0\right),\left(0,1\right),\left(0,2\right),\left(0,3\right),\left(1,0\right),\left(1,1\right),\left(1,2\right),\left(1,3\right)\right\}$

There does not exist any element in ${\mathrm{\mathbb{Z}}}_4\times {\mathrm{\mathbb{Z}}}_2$which can generate the whole group as there is no element of order $8$in ${\mathrm{\mathbb{Z}}}_4\times {\mathrm{\mathbb{Z}}}_2$. The maximum element in ${\mathrm{\mathbb{Z}}}_4\times {\mathrm{\mathbb{Z}}}_2$is $4$.

But ${\mathrm{\mathbb{Z}}}_8$is cyclic group having $4$elements of order $8,2$elements of order $4$, and$1$element of order $1$.

Thus, ${\mathrm{\mathbb{Z}}}_4\times {\mathrm{\mathbb{Z}}}_2$is not a cyclic group but ${\mathrm{\mathbb{Z}}}_8$is a cyclic group.

Therefore, ${\mathrm{\mathbb{Z}}}_4\times {\mathrm{\mathbb{Z}}}_2$is not isomorphic to ${\mathrm{\mathbb{Z}}}_8$.