Chapter 14.8, Problem 8E

a.

To determine

To make a group table for the symmetries of the square.

Answer to Problem 8E

The group table for the eight symmetries are given below,

$°$	I	${\text{R}}_j$	$R_k$	$H_0$
I	${\text{R}}_j$	${\text{R}}_j$	$R_k$	$H_0$
${\text{R}}_j$	${\text{R}}_j$	I	$H_0$	I
$R_k$	$R_k$	$H_0$	I	${\text{R}}_j$
$H_0$	$H_0$	$R_k$	${\text{R}}_j$	I

Explanation of Solution

Given information:

The figure of a square has four rotational symmetries.

The figure of square has four rotational symmetries. Each symmetry has center O and rotates the figure onto itself. Note that $90°$ rotational symmetry is another name for point symmetry.

  $\begin{array}{l}\left(1\right)\text{ 9}0°{\text{ rotational symmetry: R}}_{0,90}\\ \left(1\right)180°{\text{ rotational symmetry: R}}_{0,180}\\ \left(2\right)270°{\text{ rotational symmetry: R}}_{0,270}\\ \left(3\right)360°\text{ rotational symmetry: I is the identity}\text{.}\end{array}$

The identity mapping always maps a figure onto itself, we usually include the identity when listing the symmetries of a figure.

The group table for the eight symmetries are given below,

$°$	I	${\text{R}}_j$	$R_k$	$H_0$
I	${\text{R}}_j$	${\text{R}}_j$	$R_k$	$H_0$
${\text{R}}_j$	${\text{R}}_j$	I	$H_0$	I
$R_k$	$R_k$	$H_0$	I	${\text{R}}_j$
$H_0$	$H_0$	$R_k$	${\text{R}}_j$	I

b.

To determine

The find that the group is commutative.

Answer to Problem 8E

The group is commutative as it is an abelian group.

Explanation of Solution

Given:

The figure equilateral triangle has 4 rotational symmetry.

Concept Used:

The group is commutative if it has four property,

1. The product of two symmetry is another symmetry.
2. The set of symmetries contains the identity.
3. Each symmetry has an inverse that is also a symmetry.
4. Forming of product is an associative group,

  $A°\left(B°C\right)=\left(A°B\right)°C$ for any three group.

Therefore, the group is commutative as it is an abelian group.