UA-AUG-AUMAT229 Discovery guide

Section 2.2 Discovery guide

Group axioms.

A pair $(G, ⋆)$ consisting of a collection $G$ and a binary operation $⋆$ on $G$ is a group if the following four conditions are met.

Closure.
The collection $G$ is closed under the operation $⋆ .$ This means that when $g_{1}, g_{2}$ is a pair of objects from $G,$ then the operation result $g_{1} ⋆ g_{2}$ is always some object from $G .$
Associativity.
The operation is associative. This means that when $g_{1}, g_{2}, g_{3}$ is a triple of objects from $G,$ then the results of computing $(g_{1} ⋆ g_{2}) ⋆ g_{3}$ and $g_{1} ⋆ (g_{2} ⋆ g_{3})$ are always the same.
Identity.
The collection $G$ contains an identity/unity/neutral element, denoted $e,$ so that
$\begin{aligned} e ⋆ g & = g, & g ⋆ e = g \end{aligned}$
are both true for all objects $g$ in $G .$
Opposites.
Each object $g$ in the collection $G$ has an corresponding opposite object $\tilde{g}$ so that both
$\begin{aligned} g ⋆ \tilde{g} & = e, & \tilde{g} ⋆ g = e \end{aligned}$
are true.

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You may wish to refer back to Section 2.1 for the multiplicative and additive rewrites of these axioms.

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Discovery 2.1.

Which of the following collections are groups under the operation of addition? For those that are, make sure you know what the identity object is and what is meant by “an object's opposite.”

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(a)

The collection

Z = {\dots, - 3, - 2, - 1, 0, 1, 2, 3, \dots}

of integers.

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(b)

The collection $R$ of real numbers.

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(c)

The collection of all $m \times n$ matrices (for fixed choice of $m$ and $n$ ).

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(d)

The vector space $R^{2} .$

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Discovery 2.2.

Which of the following collections are groups under the operation of multiplication? For those that are, make sure you know what the identity object is and what is meant by “an object's opposite.”

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(a)

The collection

Z = {\dots, - 3, - 2, - 1, 0, 1, 2, 3, \dots}

of integers.

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(b)

The collection $R$ of real numbers.

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(c)

The collection of all square $n \times n$ matrices (for fixed choice of $n$ ).

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(d)

The collection of all $m \times n$ matrices (for fixed choice of $m$ and $n$ ).

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Discovery 2.3.

Some of the collections that failed in Discovery 2.2 can be “fixed” by restricting to a smaller collection. Determine which of those collections can be fixed this way, and the specific smaller collection that makes things work.

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Discovery 2.4.

Remind yourself of the axioms defining an abstract vector space ¹.

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(a)

A vector space comes equipped with two operations, addition and scalar multiplication, while a group only has one operation. If we “forget” about the operation of scalar multiplication, verify for yourself that four of the five addition axioms for a vector space are precisely the additive group axioms.

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(b)

There are five addition axioms for vector spaces but only four axioms for groups. What special property does that “extra” addition axiom confer upon vector spaces?

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Discovery 2.5.

The abstraction of the group axioms treats addition and multiplication as two examples of the same phenomenon: a binary operation that exhibits certain algebraic patterns.

But sometimes multiplication and addition really are the same thing in one example! Consider the collection $G$ of $2 \times 2$ matrices that have the form

[\begin{matrix} 1 & a \\ 0 & 1 \end{matrix}],

for all possible different values of the parameter $a .$ This collection forms a group under the binary operation of matrix multiplication.

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(a)

Compute the product

[\begin{matrix} 1 & a \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & b \\ 0 & 1 \end{matrix}] .

What is the pattern of multiplication in this group?

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(b)

What matrix acts as the identity in this group, and what is the corresponding value of the parameter $a$ for that identity object?

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(c)

Given matrix

[\begin{matrix} 1 & b \\ 0 & 1 \end{matrix}]

in this collection, where we have replaced parameter $a$ by the arbitrary value $b,$ what is its inverse? What value has replaced parameter $a$ in this inverse?

We have seen another process in our past mathematical experience that

turns multiplication into addition,
turns the additive identity $0$ into the multiplicative identity $1,$ and
turns multiplicative inversion into additive negation.

That process is exponentiation!

\begin{aligned} e^{a} e^{b} & = e^{a + b} & e^{0} & = 1 & {(e^{a})}^{- 1} = e^{- a} \end{aligned}

Could the mapping

a \mapsto [\begin{matrix} 1 & a \\ 0 & 1 \end{matrix}],

sending parameter values to $2 \times 2$ matrices of this particular form, also be some sort of “higher-dimensional exponentiation?”

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Discovery 2.6.

For the collections examined in Discovery 2.1 and Discovery 2.2 you probably just said “Yes, of course” when considering the associativity axiom. But it's important to know exactly what associativity means and how to actually check it.

Consider the twelve rotational symmetries of a regular tetrahedron, as explored in Topic 1. (Consider these as geometric transformations, not as rotation matrices. So the group operation is composition of transformations.)

Specifically, referring to the reference tetrahedron in Figure 1.0.2, consider

Rotation $R_{1}$ .
Rotation about the axis passing through the white vertex and the centroid of the opposite face, counter-clockwise (when looking through the white vertex toward the opposite face), by angle $2 π / 3 .$
Rotation $R_{2}$ .
Rotation about the axis that passes through the blue vertex and the centroid of the opposite face, counter-clockwise (when looking through the blue vertex toward the opposite face), by angle $2 π / 3 .$
Rotation $R_{3}$ .
Rotation about the axis that passes through the midpoints of the white-red and blue-green edges by angle $π .$

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(a)

The composite transformation $R_{2} R_{3}$ is the result of first performing $R_{3}$ and then $R_{2} .$ (Careful: Since this is really composition of functions, it is the right-most transformation that is applied first!) This composite is also a rotation — just as you did in the activities of Topic 1, determine the axis and angle of rotation by examining where the colour-coded vertices end up.

Hint.

Recall from your experience creating a table of rotations in Task b of Discovery 1.5 that a rotation around a vertex-face axis always fixes the vertex the axis pass through, while a a rotation around an edge-edge vertex never fixes a vertex but instead interchanges the vertices at either end of the edges that the axis passes through.

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(b)

Now take your understanding of the composite transformation $R_{2} R_{3}$ as a single rotation and perform the same sort of analysis to determine the axis and angle of rotation of the composite transformation $(R_{2} R_{3}) R_{1} .$ (Remember: In this composition, $R_{1}$ is applied first.)

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(c)

Finally, repeat the two-step analysis of Task a and Task b but with the other choice of grouping: first describe the composite $R_{3} R_{1}$ as a single rotation, then use that description to determine the composite $R_{2} (R_{3} R_{1})$ as a single rotation.

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(d)

Did the results of your geometric analyses in Task b and Task c match?

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Discovery 2.7.

Can a group contain two different objects that satisfy the identity axiom? The identity axiom just says that $G$ must contain an identity element, but doesn't restrict how many such objects there might be.

Suppose $G$ is a group with objects $e_{1}$ and $e_{2}$ that each satisfy the identity axiom. That is, both

\begin{aligned} e_{1} ⋆ g & = g, & g ⋆ e_{1} = g \end{aligned}

are true for all objects $g$ in $G,$ and so are both of

\begin{aligned} e_{2} ⋆ g & = g, & g ⋆ e_{2} = g . \end{aligned}

Using two of the four equalities above, “simplify” the expression

e_{1} ⋆ e_{2}

in two different ways to discover the answer to the question posed at the beginning of this activity.

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Discovery 2.8.

Can an object in a group have two different corresponding opposite objects? The opposites axiom just says that each object must have a corresponding opposite, but doesn't restrict how many such objects there might be.

Suppose $g$ is an object in a group $G$ so that there exist objects ${\tilde{g}}_{1}$ and ${\tilde{g}}_{2}$ that each satisfy the opposite axiom for $g .$ That is, both

\begin{aligned} g ⋆ {\tilde{g}}_{1} & = e, & {\tilde{g}}_{1} ⋆ g = e \end{aligned}

are true, and so are both of

\begin{aligned} g ⋆ {\tilde{g}}_{2} & = e, & {\tilde{g}}_{2} ⋆ g = e . \end{aligned}

Using two of the four equalities above, along with the associativity and identity axioms, “simplify” the expression

{\tilde{g}}_{1} g {\tilde{g}}_{2}

in two different ways to discover the answer to the question posed at the beginning of this activity.

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For the next two activities, make sure you understand how the opposites axiom specifies exactly how to answer the question “What is the opposite/inverse/negative of object

g ?

” — the opposite of

g

is the precise object that fills in the blanks of both of the equalities

\begin{aligned} g ⋆ & = e, & ⋆ g = e . \end{aligned}

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Discovery 2.9.

What is the opposite of the identity? As above, your formula should successfully fill in both blanks below.

\begin{aligned} e ⋆ & = e & ⋆ e = e \end{aligned}

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Discovery 2.10.

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(a)

Determine the formula for the opposite of a product $x y$ in terms of the opposites $\tilde{x}, \tilde{y}$ of the individual factors. As above, your formula should successfully fill in both blanks below.

\begin{aligned} (x y) ⋆ () & = e & () ⋆ (x y) = e \end{aligned}

Hint.

It's not $\tilde{x} \tilde{y}!$

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(b)

Interpret your formula from Task a in the multiplicative context:

{(x y)}^{- 1} = .

Hint.

Once again, the answer is not ${(x y)}^{- 1} = x^{- 1} y^{- 1}$ !

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(c)

Interpret your formula from Task a in the additive context:

- (x + y) = .

Hint.

Unfortunately, the answer is neither

- (x + y) = - x - y

nor

- (x + y) = (- x) + (- y) .

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As usual, for positive integer

n

and element

g

in a multiplicative group

G,

we write

g^{n}

to mean

g g \dots g,

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the product of

n

factors of element

g .

n

is negative, then we take

g^{n}

to mean the product of

| n |

factors of the inverse of

g .

By convention, we also take

g^{0}

to always mean

e,

the multiplicative identity in

G .

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In an additive group, we use similar additive notation: for positive

n

write

n \cdot g

to mean

g + g + \dots + g,

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the sum of

n

terms of element

g .

n

is negative, take

n \cdot g

to mean the sum of

| n |

terms of the negative (i.e. additive inverse) of

g .

And we take

0 \cdot g

to always mean

0 .

(Careful: there are two different zeros here — one is the integer

0

and one is the additive identity in the group

G,

and they are not the same!)

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Discovery 2.11.

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(a)

Suppose $g$ is an element of a multiplicative group $G .$ What does the pattern from Task b of Discovery 2.10 say about the inverse of $g^{2} ?$ Extrapolate to a statement about the inverse of $g^{n} .$

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(b)

Suppose $g$ is an element of an additive group $G .$ What does the pattern from Task c of Discovery 2.10 say about the negative of $2 \cdot g ?$ Extrapolate to a statement about the negative of $n \cdot g .$

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One method of analyzing and keeping track of the rotational symmetries of the tetrahedron from Topic 1 is to determine how the rotation moves around the colour-coded vertices. If we focus only on the change in positions of the colour-coded vertices and ignore the geometric process of rotation that brought about this change in positions, then essentially we are considering each rotational symmetry to be a process of “mixing up” or permuting the vertices. The next activity establishes a basic property of groups that will eventually let us conclude that all groups are somehow collections of permutations.

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Discovery 2.12.

Suppose $x, y$ are objects in a group $G .$ Demonstrate that there exist objects $w, z$ in $G$ (by obtaining formulas for $w$ and $z$ in terms of $x$ and $y$ ) so that both

\begin{aligned} w x & = y, & x z & = y \end{aligned}

are true.

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