Grade 9 & 10 Introduction to Group Theory Part 1 Problem Set

In Example 4 we saw that $(Z, +)$ is a group. Now consider subtraction $-$ on $Z$ . Convince yourself that subtraction is a binary operation on $Z$ . Show that subtraction on $Z$ is not associative and use this to conclude that $(Z, -)$ is not a group.

Solution:

We know that if $a, b \in Z$ , then $a - b \in Z$ . So, subtraction is a binary operation on $Z$ . Next, let’s show that subtraction on $Z$ is not associative. To do this, we need to find integers $a, b, c \in Z$ such that $a - (b - c) \neq (a - b) - c .$ Here the symbol $\neq$ means "not equal". If $a = 1, b = 2, c = 3$ , then $a - (b - c) = 1 - (2 - 3) = 1 - (- 1) = 1 + 1 = 0$ and $(a - b) - c = (1 - 2) - 3 = (- 1) - 3 = - 1 - 3 = - 4.$ But $0 \neq - 4$ , so $1 - (2 - 3) \neq (1 - 2) - 3.$ This shows that subtraction on $Z$ is not associative, and so Axiom 1 of the group axioms is not satisfied. So $(Z, -)$ is not a group.
Note: It is true that $a - (b - c) \neq (a - b) - c$ for any $a, b, c \in Z$ as long as $c \neq 0$ . Can you show this?

Consider the set of natural numbers $N = {0, 1, 2, 3, . . .}$ , which is the set containing all non-negative whole numbers. Addition is a binary operation on $N$ . Show that $(N, +)$ is not a group.

Solution:

To show that $(N, +)$ is not a group, we need to show that at least one of the 3 group axioms does not hold. Axiom 1 holds because it doesn’t matter what order we add numbers in. We see that Axiom 2 holds because $0 + n = n = n + 0$ for any $n \in N$ , and so $0$ is the identity element for $N$ with addition.
We claim that Axiom 3 does not hold. To show that Axiom 3 does not hold, we need to find at least one element of $N$ that does not have an inverse. It turns out that every element of $N$ , aside from $0$ , does not have an inverse. To see this, let $n \in N$ and assume that $n$ is not zero. If $n$ had an inverse, say $n^{'}$ , then we would have that $n + n^{'} = 0$ and so $n^{'} = - n$ . So, if $n$ had an inverse then it would have to be $- n$ . However, since $n$ is not zero, we have that $n > 0$ . This implies that $- n < 0$ , which means that $- n$ is a negative integer, and so $- n \notin N$ . But remember that by definition of inverse, if $n$ had an inverse it would have to be an element of $N$ . So, this shows that $n$ does not have an inverse.

Consider the set of even integers $2 Z = {\dots, - 6, - 4, - 2, 0, 2, 4, 6, \dots} .$ Convince yourself that addition is a binary operation on $2 Z$ . Show that $(2 Z, +)$ is a group.

Solution:

To convince ourselves that addition is a binary operation on $2 Z$ , let’s check that the addition of any two elements of $2 Z$ is again an element of $2 Z$ . So, let $x, y \in 2 Z$ . We know that the addition of two integers is again an integer, so $x + y$ is an integer. It remains to show that $x + y$ is even. Since $x$ and $y$ are even, we can write $x = 2 k and y = 2 ℓ$ for some $k, ℓ \in Z$ . Then $x + y = 2 k + 2 ℓ = 2 (k + ℓ),$ which shows that $x + y$ is even. We conclude that $x + y \in 2 Z$ , and so addition is a binary operation on $2 Z$ . To show that $(2 Z, +)$ is a group we need to check that the 3 group axioms hold. Let’s go through each axiom:

Axiom 1: It doesn’t matter what order we add numbers in. So, associativity holds.

Axiom 2: The identity element is $0$ because the addition of $0$ with any number is that number again. In particular, for any $x \in 2 Z$ , $0 + x = x = x + 0.$

Axiom 3: We need to show that every element of $2 Z$ has an inverse. So, let $x \in 2 Z$ . Note that $- x \in 2 Z$ because the negative of an even integer is still an even integer. Then $(- x) + x = x + (- x) = x - x = 0.$ This shows that $- x$ is the inverse of $x$ . So, every element in $2 Z$ has an inverse.

Suppose we are given a group $(G, ∙)$ . Axiom 2 of the group axioms says that $(G, ∙)$ has an identity element, which we denote by ${id}_{G} \in G$ . Show that the identity element ${id}_{G}$ is unique. In other words, show that there is exactly one element of $G$ that satisfies the property: for every $a \in G$ , $a ∙ {id}_{G} = a = {id}_{G} ∙ a$ .

Hint: If $e \in G$ satisfies the above property, try to show that $e = {id}_{G}$ .

Solution:

Suppose there exists $e \in G$ such that for every $a \in G$ , $a ∙ e = a = e ∙ a .$ Since this equation holds for every element $a \in G$ , it holds for ${id}_{G}$ . So ${id}_{G} ∙ e = {id}_{G} .$ We know ${id}_{G}$ satisfies the same property as $e$ . That is, for every $a \in G$ , $a ∙ {id}_{G} = a = {id}_{G} ∙ a .$ Since this equation holds for every element $a \in G$ , it holds for $e$ . So $e = {id}_{G} ∙ e .$ Together, Equations (1) and (2) tell us that $e = {id}_{G} ∙ e = {id}_{G} .$ This shows that ${id}_{G}$ is the only element of the group that satisfies the above property.

Consider the following triangle:

An isosceles triangle with its smallest side as a horizontal base and longer equal slanted sides.

Write down the symmetries of this triangle and compare them with the symmetries of the equilateral triangle from Exercise 4.

Solution:

The complete list of symmetries of the above triangle are as follows:

Identity symmetry and reflection in the axis through the top vertex and midpoint of the base.

In words, these symmetries are the "do nothing symmetry" (which is the same as rotation by $360$ degrees) and reflection in the axis that goes through the center and top tip of the triangle. The symmetries of this triangle are symmetries of the equilateral triangle in Exercise 4. However, the equilateral triangle in Exercise 4 has way more symmetries than this triangle. So, the sets of symmetries are different.

Consider the following hexagon:

Write down the symmetries of this hexagon. Then compare the symmetries of this hexagon with the symmetries of the benzene molecule, which were found in Exercise 6. Are they the same in some sense? Or are they different?

Solution:

The complete list of symmetries of a hexagon are as follows:

Identity symmetry; 5 counterclockwise rotations by 60, 120, 180, 240 and 300 degrees; 3 reflections in axes through the midpoints of two opposite sides of the hexagon; 3 reflections in axes through two opposite vertices of the hexagon.

There are $6$ counter clockwise rotations by multiples of $60$ degrees. And there are $6$ reflections in the axes shown above. If you compare these $12$ symmetries, with the $12$ symmetries of the benzene molecule, then we see that they are exactly the same! This is because the benzene molecule has a similar structure to that of the hexagon. You can think of the benzene molecule as a hexagon but with 6 arms sticking out, and these arms do not affect the symmetries.

Consider the following shapes.

Eight shapes: equilateral triangle, square, regular pentagon, regular hexagon, and also regular polygons with 7, 8, 9 and 10 sides.

These shapes are examples of regular polygons. A polygon is called regular if all of its sides have the same length and all of its angels are the same. For example, an equilateral triangle is a regular polygon. Write down the symmetry group of an arbitrary regular polygon.

Hint: Let $P_{n}$ be a regular polygon with $n$ sides. The goal is to write down the elements of $Sym (P_{n})$ . In Exercise 4 you computed $Sym (P_{3})$ , and in Problem 6 you computed $Sym (P_{6})$ . If you compute $Sym (P_{n})$ for small $n$ , say $n \in {3, 4, 5, 6}$ , do you see a pattern?

Solution:

Let $P_{n}$ be a regular polygon with $n$ sides. We want to find the elements of $Sym (P_{n})$ .
In Exercise 4, we saw that $Sym (P_{3})$ consists of 3 counter clockwise rotations and 3 reflections. The rotations are by multiplies of $120 = 360 / 3$ . And for each tip of the triangle, there is an axis of reflection that goes through the tip and the center of the triangle.
Next, consider the square $P_{4}$ :

The square has 4 rotations that are symmetries. These are counter clockwise rotations by $0$ (or $360$ ), $90$ , $180$ , and $270$ degrees. There are $4$ reflections of the square which are symmetries. These are depicted in the following photo:

Four lines all passing through the centre of the square. Two lines lie along the diagonals of the square and two lines pass through the midpoints of opposite sides of the square.

Reflection in each of these $4$ axes is a symmetry. To conclude, $Sym (P_{4})$ consists of $4$ rotations which are by multiplies of $90 = \frac{360}{4}$ degrees and $4$ reflections.
For another example, consider the pentagon $P_{5}$ :

Five lines all passing through the centre of the pentagon. Each of the five lines passes through one of the vertices of the pentagon and the midpoint of the opposite side.

The pentagon has $5$ rotations that are symmetries. These are counter clockwise rotations by $0$ (or $360$ ), $72$ , $144$ , $216$ , and $288$ degrees. There are $5$ reflections of the pentagon which are symmetries. These are depicted in the following photo:

Reflection in each of these $5$ axes is a symmetry. To conclude, $Sym (P_{5})$ consists of $5$ rotations which are by multiplies of $72 = 360 / 5$ degrees and $5$ reflections.
Given these $3$ examples, we expect that $Sym (P_{n})$ consists of $n$ counter clockwise rotations by multiples of $\frac{360}{n}$ degrees and $n$ reflections. And indeed, this is true. We do not have the tools to prove this yet, but hopefully the above examples convince you that it is a reasonable expectation.