Question

List all subgroups of${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$ . (There are more than two.)

Short answer

All subgroup of${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$ are$\left\{\left(0,0\right)\right\};\left\{\left(0,0\right),\left(1,0\right)\right\};\left\{\left(0,0\right)\left(0,1\right)\right\};\left\{\left(0,0\right),\left(1,1\right)\right\}$ .

Step-by-step solution

Subgroups of a group

If Aand B are normal subgroups of a group G such that $G=AB$and , then $G=A\times B$.

Find all subgroups of ℤ2×ℤ2

Consider the field ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$.

It is known that,${\mathrm{\mathbb{Z}}}_2=\left\{0,1\right\}$.

The element in ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$is as follows:

$\begin{array}{rcl}{\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2& =& \left\{0,1\right\}\times \left\{0,1\right\}\\ & =& \left\{\left(0,0\right),\left(0,1\right),\left(1,0\right),\left(1,1\right)\right\}\end{array}$

Since the subgroups is non-trivial, so the subgroups have to order less than 1.

Also, the subgroups are proper, so the order of the subgroups must be less than 4 .

It can be noted that every order of subgroup has to divide the order of ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$, that is 4 .

So, only proper subgroup of ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$is of order 2 .

Also, identity (0, 0) of ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$is included in subgroup.

Find remaining elements in the subgroup as follows:

Consider element (0, 1) and find their addition as:

(0, 1) + (0, 1) = (0 + 0, 1 + 1)

= (0, 2)

In all elements ${\mathrm{\mathbb{Z}}}_2$, $0\equiv 2$mod2 we have:

(0, 1) + (0, 1) = (0, 0)

Hence, (0, 1) has order 2 .

Thus,$\left\{\left(0,0\right),\left(0,1\right)\right\}$ is a subgroup of${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$ .

Consider element (1, 0) and find their addition as:

$\begin{array}{rcl}\left(1,0\right)+\left(1,0\right)& =& \left(1+1,0+0\right)\\ & =& \left(2,0\right)\end{array}$

In all elements ${\mathrm{\mathbb{Z}}}_2,0\equiv 2$mod2 we have:

$\left(1,0\right)+\left(1,0\right)=\left(0,0\right)$

Therefore, (1, 0) has order 2 .

Hence,$\left\{\left(0,0\right),\left(1,0\right)\right\}$ is a subgroup of${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$ .

Consider element (1, 1) and find their addition as:

$\begin{array}{rcl}\left(1,1\right)+\left(1,1\right)& =& \left(1+1,1+1\right)\\ & =& \left(2,2\right)\end{array}$

In all elements ${\mathrm{\mathbb{Z}}}_2,0\equiv 2$mod2we have:

$\left(1,1\right)+\left(1,1\right)=\left(0,0\right)$

Therefore, (1, 1) has order 2 .

Hence,$\left\{\left(0,0\right),\left(1,1\right)\right\}$ is a subgroup of ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$.

Therefore, all subgroup of ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$are

$\left\{\left(0,0\right)\right\};\left\{\left(0,0\right),\left(1,0\right)\right\};\left\{\left(0,0\right)\left(0,1\right)\right\};\left\{\left(0,0\right),\left(1,1\right)\right\}$