Consider the set \(\{-1,1\}\). Convince yourself that multiplication \(\times\) is a binary operation on \(\{-1,1\}\). Show that \((\{-1,1\},\times)\) is a group.
To convince ourselves that multiplication is a binary operation on \(\{-1,1\}\), let’s check that the multiplication of any two elements of \(\{-1,1\}\) is again an element of \(\{-1,1\}\). Indeed, this is true since: \[(-1)\times (-1) = 1, \quad (-1)\times 1 = -1,\quad 1\times (-1) = -1, \quad 1\times 1=1.\] To show that \((\{-1,1\},\times)\) 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 multiply numbers in. So, associativity holds.
Axiom 2: The identity element is \(1\) because the multiplication of \(1\) with any number is that number again. In particular, \[1\times 1 = 1, \quad 1\times (-1) = -1 = (-1)\times 1.\]
Axiom 3: The inverse of \(-1\) is \(-1\) since \((-1)\times (-1) = 1\). The inverse of \(1\) is \(1\) since \(1\times 1 = 1\). This shows that every element in \(\{-1,1\}\) has an inverse.
In Example 4 we saw that \((\mathbb{Z},+)\) is a group. Now consider multiplication \(\times\) on \(\mathbb{Z}\). Convince yourself that \(\times\) is a binary operation on \(\mathbb{Z}\). Is \((\mathbb{Z},\times)\) a group?
We know that if \(a\in\mathbb{Z}\)
and \(b\in\mathbb{Z}\), then \(a\times b= ab\in\mathbb{Z}\). So,
multiplication is a binary operation on \(\mathbb{Z}\). Although multiplication is a
binary operation on \(\mathbb{Z}\),
\((\mathbb{Z},\times)\) is
not a group. To show that \((\mathbb{Z},\times)\) is not a group, we
need to show that at least one of the 3 group axioms does not hold.
Axiom 1 is satisfied because it doesn’t matter what order we multiply
numbers in. And Axiom 2 is satisfied because \(1\) is the identity element.
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 \(\mathbb{Z}\) that does not have an inverse.
The element \(0\in\mathbb{Z}\) does not
have an inverse because \(0\) times any
number is \(0\), not 1. That is, for
any \(a\in\mathbb{Z}\), \[0\times a = 0 = a\times 0\] which is not
the identity element \(1\). So, \(0\) does not have an inverse. This shows
that Axiom 3 does not hold and hence \((\mathbb{Z},\times)\) is not a group. In
fact, the only element in \(\mathbb{Z}\) that has an inverse is \(1\). Can you show that any \(a\in\mathbb{Z}\) satisfying \(a\not\in\{0,1\}\) does not have an
inverse?
Recall that a rational number is of the form \(\frac{a}{b}\) where \(a,b\in\mathbb{Z}\) and \(b\) is not zero. Let \(\mathbb{Q}\) be the set of all rational numbers. And let \(\mathbb{Q}^\ast\) be the set \(\mathbb{Q}\) but with \(0\) removed. Recall that we multiply two rational numbers by \[\frac{a}{b}\times \frac{a'}{b'} = \frac{aa'}{bb'}.\] Convince yourself that multiplication \(\times\) is a binary operation on \(\mathbb{Q}^\ast\). Is \((\mathbb{Q}^\ast,\times)\) a group?
Let \(a/b\in\mathbb{Q}^\ast\) and \(a'/b'\in\mathbb{Q}^\ast\). By definition of \(\mathbb{Q}^\ast\), \(b\) and \(b'\) are not zero and so \(bb'\) is not zero. Thus, the multiplication \[\frac{a}{b}\times\frac{a'}{b'} = \frac{aa'}{bb'}\] is in \(\mathbb{Q}\) since \(aa',bb'\in\mathbb{Z}\) and \(bb'\) is not zero. It remains to show that the multiplication of \(a/b\) with \(a'/b'\) is not zero. Well, since these elements are not zero, \(a\) and \(a'\) are not zero. Then \(aa'\) is not zero, and so \(aa'/bb'\) is not zero. So, \(aa'/bb'\in\mathbb{Q}^\ast\). We conclude that multiplication is a binary operation on \(\mathbb{Q}^\ast\). We claim that \((\mathbb{Q}^\ast,\times)\) is a group. Let’s go through each group axiom:
Axiom 1: It doesn’t matter what order we multiply numbers in. So, associativity holds.
Axiom 2: The identity element is \(1=1/1\) because the multiplication of \(1\) with any number is that number again. In particular, for any \(a/b\in\mathbb{Q}^\ast\), we have that \[1\times \frac{a}{b} = \frac{a}{b} = \frac{a}{b}\times 1.\]
Axiom 3: We need to show that every element in \(\mathbb{Q}^\ast\) has an inverse. So, let \(a/b\in\mathbb{Q}^\ast\). Because \(a\) and \(b\) are not zero, the element \(b/a\) is also in \(\mathbb{Q}^\ast\). We compute that \[\frac{a}{b} \times \frac{b}{a} = \frac{ab}{ba} = \frac{ab}{ab} = 1,\] and similarly \[\frac{b}{a}\times\frac{a}{b}=1.\] This shows that \(b/a\) is the inverse of \(a/b\). So, every element in \(\mathbb{Q}^\ast\) has an inverse.
An equilateral triangle is a triangle whose 3 sides all have the same length. The triangle in Example 6 is an equilateral triangle. Write down all of the symmetries of an equilateral triangle
The complete list of symmetries of an equilateral triangle are as follows:
In words, these symmetries are:
The "do nothing symmetry", which is the same as rotation by \(360\) degrees. This is labelled as the "identity symmetry" because it is the identity element of the group \((\text{Sym}(T),\circ)\), where \(T\) is an equilateral triangle (we will see this in Exercise 5).
Counter clockwise rotation by \(120\) degrees
Counter clockwise rotation by \(240\) degress.
Reflection in the axis that goes through the center and top tip of the triangle.
Reflection in the axis that goes through the center and left tip of the triangle.
Reflection in the axis that goes through the center and right tip of the triangle.
Note that clockwise rotations by multiples of \(120\) degrees are also symmetries. However, these symmetries are already in the above list. For example, rotation clockwise by \(120\) degrees is the same as counter clockwise rotation by \(240\) degrees. If you prefer, you can replace "counter clockwise" with "clockwise" in the above list, it doesn’t matter at all. Also note that rotations by \(n120\) degrees with \(n\geq 4\) are symmetries. Again, they are already in the above list. For example, counter clockwise rotation by \(480\) degrees is the same as counter clockwise rotation by \(120\) degrees.
Let \(T\) be an equilateral triangle. In Exercise 4 you computed \(\text{Sym}(T)\). Convince yourself that \((\text{Sym}(T),\circ)\) is a group.
We need to convince ourselves that the 3 group axioms hold for \(\text{Sym}(T)\) with composition. Let’s go through each axiom:
Axiom 1: For all \(P,Q,R\in\text{Sym}(T)\), we need \((P\circ Q)\circ R\) to be the same symmetry as \(P\circ(Q\circ R)\). Proving this is a bit tough given our current tools, so let’s do a concrete example to see why it works.
Let \(P\) be counter clockwise rotation by \(120\) degrees, let \(Q\) be reflection in the axis that goes through the center and top tip of the triangle, and let \(R\) be counter clockwise rotation by \(240\) degrees. These symmetries are illustrated as follows:
The compositions \(P\circ Q\) and \(Q\circ R\) are computed to be:
We see that these two compositions are the same symmetry, and they are equal to the symmetry that reflects in the axis going through the center and right tip of the triangle. Using these compositions, we compute \(P\circ (Q\circ R)\) and \((P\circ Q)\circ R\):
We see that \(P\circ(Q\circ R) = (P\circ Q)\circ R\), as desired. They are equal to the symmetry that reflects in the axis going through the center and left tip of the triangle:
Axiom 2: The identity element in \(\text{Sym}(T)\) is the "do nothing" symmetry. Here is an illustration of the identity element:
Let \(P\) be any symmetry of \(T\). If you do nothing to \(T\) and then apply \(P\), it’s the same as just applying \(P\) to \(T\). Similarly, if you apply \(P\) to \(T\) and then do nothing, it’s the same as just applying \(P\) to \(T\). So indeed, this is the identity element \(\text{id}_{\text{Sym}(T)}\).
Let’s do a concrete example where \(P\) is chosen to be counter clockwise rotation by \(120\) degrees. Here is the composition \(P\circ\text{id}_{\text{Sym}(T)}\):
We see that this composition is just \(P\). Here is the composition \(\text{id}_{\text{Sym}(T)}\circ P\):
Again, we see that this composition is just \(P\).
Axiom 3: We need to show that every element in \(\text{Sym}(T)\) has an inverse. For this, it will be helpful to recall the elements of \(\text{Sym}(T)\) from Exercise 4.
First consider the symmetries that are reflections. The inverse of a reflection is itself. In other words, if \(R\in\text{Sym}(T)\) is a reflection then the inverse of \(R\) is \(R\), that is \(R^{-1}=R\). This is because if we reflect in the same axis twice, it’s the same as doing nothing. Here is a concrete example:
We see that this composition is just \(\text{id}_{\text{Sym}(T)}\). Next, consider symmetries that are rotations. If \(S\in\text{Sym}(T)\) is a rotation, then by Exercise 4, it’s counter clockwise rotation by \(n120\) degrees for some \(n\in\{0,1,2\}\). The inverse of counter clockwise rotation by \(n120\) degrees is the symmetry that reverses this rotation, which is clockwise rotation by \(n120\) degrees. Note that clockwise rotation by \(n120\) degrees is the same symmetry has counter clockwise rotation by \((360 - n120)\) degrees. Here is a concrete example where we take \(n=1\):
We see that these two compositions are just \(\text{id}_{\text{Sym}(T)}\).
Consider the benzene molecule below and denote it by \(BM\). The benzene molecule is a shape, and so \((\text{Sym}(BM),\circ)\) is a group. Chemists study the symmetries of molecules such as \(BM\), and classify molecules according to their symmetry group. The study of such symmetries can be used to predict or explain chemical properties of a molecule.
Write down the symmetry group of the benzene molecule \(BM\). In other words, write down the elements of \(\text{Sym}(BM)\).
The complete list of symmetries of the benzene molecule are as follows:
