Note that, were we to rotate the square, there is a maximum of four (4) different positions in which it would occupy the space it occupies now, and a total of two (2) possible orientations – face up and face down.
This means that there are a total of eight (8) different positions (4 × 2 = 8) in which the square will fit in the space it originally occupied.
Now, to form a group we need to create a nonempty set that has elements which can be “paired-up” in order to get yet another element of the set.
In our set, those elements consist of the ways in which we can manipulate the transparent square so that it occupies its original position.
Let’s assign the following codes to the following manipulations:
CODE
Any of the eight positions previously mentioned can be obtained by performing one of the above manipulations. Moreover, we can perform a sequence of two manipulations and get the same result as a single manipulation all by itself. Consequently, we are now able to form a group.
Suppose that the square flipped 180° about its horizontal axis and then rotated 90° counter clockwise as follows . . .