Question

Show that the additive group $Z_2\times Z_4$is not cyclic but is generated by two elements.

Short answer

It is proved that_$Z_2\times Z_4$_is not cyclic but generated by two elements.

Step-by-step solution

Definition of a cyclic group

A cyclic group G is a group in which there is at least one element such as $a\in G$so that {a} = {${\mathrm{a}}^{\mathrm{n}}|\mathrm{a}\in \mathrm{Z}$}is a subgroup of G.

Proving that Z2 x Z4 is not cyclic.

The two groups are:

$$\begin{gathered} {\mathrm{Z}}_2=\{0,1\} \\ {\mathrm{Z}}_4=\{0,1,2,3\} \end{gathered}$$

Thus Z₂ x Z₄is:

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

For_Z₂ x Z₄ to be cyclic, there should be at least one element that can be a generator of this entire group.

Let us analyze the cyclic groups generated by each of these elements.

<(0,0)> = {(0,0)} has order 1.

<(1,0)> = {(1,0), (0,0)} has order 2.

<(0,1)> = {(0,1),(0,0)} has order 2.

<(1,1)> = {(1,1), (0,2), (1,3), (0,0)} has order 4.

<(0,2)> = {(0,2), (0,3), (0,0), (0,1)} has order 4.

<(1,2)> = {(1,2), (0,3), (1,0), (0,1)} has order 4.

<(0,3)> = {(0,3), (0,2), (0,0), (0,1)} has order 4.

<(1,3)> = {(1,3), (0,2), (0,0), (0,1)} has order 4.

As seen above, none of the elements can generate the entire group_Z₂ x Z₄ . Hence, the group Z₂ x Z_{4_}is not cyclic.

Proving that Z2 x Z4can be generated by two elements.

As we see that the order of group Z₂ x Z₄ is 8. This implies that we will need two unique elements, each having order 4, to generate the entire group Z₂ x Z₄. One such example can be the combination of elements (1,1) and (0,2). Hence, group Z₂ x Z₄ can be generated by two elements.