Imagine a game board with four squares (like a \(1 \times 4\) chessboard) labelled \(0\) through \(3\text{,}\) and one token placed on a square. This is a rather simple game, all we can do is move the token to the right, and if the token moves off the board, it goes back to spot \(0\text{.}\)
We can map out all possible moves of this game with a simple chart. We’ll write \(a + b\) to mean the token starts in slot \(a\) and travels \(b\) moves to the right.
Activity4.2.1.
Fill out the chart of all possible moves of the token on a \(1 \times 4\) chessboard.
Table4.2.1.
\(a \backslash b\)
\(0\)
\(1\)
\(2\)
\(3\)
\(0\)
\(\)
\(\)
\(\)
\(\)
\(1\)
\(\)
\(\)
\(\)
\(\)
\(2\)
\(\)
\(\)
\(\)
\(\)
\(3\)
\(\)
\(\)
\(\)
\(\)
This game is rather boring... is it even a game? But it illustrates how we can uncover the underlying structure mathematically of something.
Subsection4.2.2Algebra’s Goal
If we ever make contact with an extraterrestrial society, how could we ever hope to communicate with them? We think that mathematics is our best chance of communicating with intelligent life. We are pretty confident that if they have the means to send and receive messages, they will have an understanding of mathematics.
But what does their alien math look like? We would be very suprised to find the Mean Value Theorem or the Four Color Theorem in their writings, right? At least with the same phrasing we use, BUT we would not be surprised if they had their own versions of these results.
Modern algebra seeks to explore "deeper" structures of mathematical systems, so that we can recognize familiar math in unfamiliar environments. Can we find the big picture patterns of mathematics? That is our goal with higher algebra.
Subsection4.2.3A Familiar Pattern
Let’s find another underlying structure. Define the set \(A = \{1,i,-1,-i\}\) where \(i = \sqrt{-1}\text{,}\) and consider multiplication in this set. We should point out that if we multiply two things in \(A\text{,}\) the result is also in \(A\text{,}\) and this property is called closure. As an example, \(i \cdot i = i^2 = (\sqrt{-1})^2 = -1 \in A. \)
Activity4.2.2.
Fill out the multiplication table for set \(A = \{1,i,-1,-i\}.\)
Table4.2.2.
\(\cdot\)
\(1\)
\(i\)
\(-1\)
\(-i\)
\(\; 1\)
\(\)
\(\)
\(\)
\(\)
\(\; i\)
\(\)
\(-1\)
\(\)
\(\)
\(-1\)
\(\)
\(\)
\(\)
\(\)
\(-i\)
\(\)
\(\)
\(\)
\(\)
Do you notice any similarities between Table 4.2.1 and ? Tables like these are called Cayley Tables, and one use of Cayley Tables is highlighting patterns. Even though we have different contexts, different sets, and even different operations, in some way they are the same thing.
Figure4.2.3.Underlying Cayley Table
Algebraists can show that the context of the token on the chessboard and the multiplicative powers of \(i\) are, in fact, the same. What we mean, actually, is that they operate the same way. We say they are isomorphic (meaning "same shape"), and in an algebra course we can prove these things are isomorphic.
Subsection4.2.4What is a Group?
Subsubsection4.2.4.1Introduction
Both of the previous examples are examles of one of the simplest algebraic structures: the group. We have found two groups, so far, and highlighted that they are isomorphic, but what exactly is a group?
The group \(G\) is a set paired with an operation (let’s say \(\ast\)) that satisfies a short list of properties. The first two are familar to us.
the closure property: if \(a, b \in G\text{,}\) then \(a \ast b \in G.\)
the associative property: for any \(a,b,c \in G\text{,}\)\((a \ast b) \ast c = a \ast (b \ast c)\)
To show the last two properties, let’s consider the two examples we discussed earlier. Let \(A\) be paired with multiplication as earlier, and let \(B = \{0,1,2,3\}\) be paired with addition defined by such: the token starting at square \(a\) moving \(b\) squares to the right will rest on square \(a + b = c\text{.}\)
Subsubsection4.2.4.2Identity Elements
How is a number affected when multiplied by \(1\text{?}\) What about when summed with \(0\text{?}\)
By using the Caylet Tables from earlier, we can see in set \(A\text{,}\) there is an element, \(e\text{,}\) so that for any other element \(a \in A\text{,}\)\(a \cdot e = e \cdot a = a\text{.}\) In other words, multiplying by \(e\) changes nothing. An element like \(e\) is known as the identity of \(G\text{.}\)
Question4.2.4.
What is the identity of the group \(A\) with operation \(\cdot\text{?}\)
Solution.
The identity is \(1\text{.}\)
Question4.2.5.
What is the identity of the group \(B\) with operation \(+\text{?}\)
Solution.
The identity is \(0\text{.}\)
The third property of groups is that every group has an identity. There is an element \(e\) such for any element \(g \in G\text{,}\)\(g \ast e = e\ast g = g\text{.}\)
Subsubsection4.2.4.3Inverse Elements
Consider the token game once again, and suppose we start on square \(0\) and move \(3\) spaces to right to end up on square \(3\text{.}\) Can we make another move to end up back on square \(0\text{?}\) In general, if we start on square \(0\) and move \(b\) squares to the right. We can only move right, but is it possible to keep moving right until we end up back on square \(0\text{?}\)
Question4.2.6.
After moving \(a\) squares, starting at square \(0\text{,}\) determine how many squares you must traverse to get back to square \(0\) for \(a \in \{0,1,2,3\}\text{.}\)
Solution.
If \(a = 0\text{,}\) we must move \(0\) squares. If \(a = 1\text{,}\) we must traverse \(3\) squares to return to square \(0\text{.}\) Similarly, we travel \(2\) squares and \(1\) square respectively for \(a = 2, 3\) respectively.
So for any square \(a\text{,}\) there is a number of squares \(b\) that results in landing on square \(0\text{.}\) In other words, for any \(a \in B\text{,}\) there is a \(b \in B\) where \(a + b = 0\text{.}\) In this case, we say \(b\) is the inverse of \(a\text{.}\) Remember that \(0\) serves a special role in \(B\text{:}\) the identity.
The last property of group is the inverse property. That is, for any \(g \in G\text{,}\) there is an element \(h \in G\text{,}\) where \(g \ast h = h \ast g = e\text{,}\) where \(e\) is the identity of \(G\text{.}\) In other words, starting with any element, we can get back to the group’s identity through the inverse. We often denote the inverse of \(g\) as \(g^{-1}\text{.}\)
Question4.2.7.
Determine the inverses of the elements of \(A\text{.}\)
Solution.
We know \(1 \cdot 1 = 1\text{,}\) so \(1\) is its own inverse. Also \(i \cdot -i = -i \cdot i = 1\text{,}\) and so \(i\) and \(-i\) are inverses of each other. Lastly, \(-1 \cdot -1 = 1\text{,}\) and so \(-1\) is its own inverse.
Subsection4.2.5Why do we think about groups?
If we can prove a statement about a generic group \(G\text{,}\) then we know that statement will apply to any group we find! Knowing things about groups will help us find the familiar patterns in unfamiliar environments.
So proving a statement about a group \(G\) proves statements in an many different contexts (sometimes an infinite number of contexts). Thinking in generalized terms is a powerful way to approach math, and mathematicians like to generalize for that very reason.
For example, here are some things we know about groups.
Theorem4.2.8.
Let \(G\) be a group. Then
\(G\) has exactly one identity.
for any \(a \in G\text{,}\)\(a\) has exactly one inverse.
for any \(a \in G\text{,}\)\((a^{-1})^{-1} = a\text{.}\)
for any \(a, b \in G\text{,}\)\((ab)^{-1} = b^{-1}a^{-1}.\)
You might know some groups already. Do you think the following sets are groups with the given operation?
Question4.2.9.
Is set \(A\) with operation \(\ast\) a group? If not, why? For our purposes, just think about the identity and inverses.
\(A:\) all integers; \(\ast:\) addition
\(A:\) all real numbers; \(\ast:\) addition
\(A:\) all real numbers; \(\ast:\) multiplication
\(A:\) all real numbers except \(0\text{;}\)\(\ast:\) multiplication
Solution.
\(A:\) all integers; \(\ast:\) addition
Yes! The identity is \(0\text{,}\) and the inverse of \(a\) is \(-a\text{.}\)
\(A:\) all real numbers; \(\ast:\) addition
Yes! The identity is \(0\text{,}\) and the inverse of \(a\) is \(-a\text{.}\)
\(A:\) all real numbers; \(\ast:\) multiplication
No! :( The identity is \(1\text{,}\) but \(0 \cdot a \neq 1\) for any \(a\text{.}\) So \(0\) has no inverse.
\(A:\) all real numbers except \(0\text{;}\)\(\ast:\) multiplication
Yes! The identity is \(1\text{,}\) and the inverse of \(a\) is \(1/a\text{.}\) If \(0\) is not in the picture, we can never obtain \(0\) through multiplication.
Subsection4.2.6A Group of Squares
Let’s look at one last example of a group. Take a square with corners labelled \(a, b, c,d\) (We would recommend cutting one out of paper), and we will say a "symmetry" of the square is an action in which we pick up the square, do something to it, and put it down again so that it takes up the same space it did before. For example, we can rotate the square \(90^{\circ}\text{,}\) and it will rest on the exact same space where it started.
There are eight symmetries of the square. Can you find them?
Question4.2.10.
What are the eight basic symmetries of the square?
Solution.
There are four rotations: rotating \(0^{\circ}\text{,}\)\(90^{\circ}\text{,}\)\(180^{\circ}\text{,}\)\(270^{\circ}\text{.}\)
There are also four flips: flip along a horizontal axis, along a vertical axis, along a northwest diagonal, and along a northeast diagonal.
Let’s call this set of symmetries \(D = \{I, R_1, R_2, R_3, F_h, F_v, F_{NW}, F_{NE}\}\text{.}\) Here, we call a rotation of \(0^{\circ}\)\(I\text{.}\)
FIGURE
What if we did two of these symmetries? Do we get something new? For symmetries \(\alpha\) and \(\beta\text{,}\) the composition \(\alpha \circ \beta\) is the result of first applying \(\beta\) and then applying \(\alpha\text{.}\)
For example, \(R_1 \circ R_2\) means we first rotate the square \(180^{\circ}\) and then another \(90^{\circ}\text{.}\) Is this a new symmetry? No... it’s just \(R_3\text{.}\) Are any of them new?
Activity4.2.3.
In groups, fill out a Cayley table for \(D\text{.}\) Don’t assumer that \(\alpha \circ \beta =
\beta \circ \alpha.\)Figure 4.2.12 shows a calculator for the symmetries of a squre. This may be useful!
Table4.2.11.
\(\circ\)
\(I\)
\(R_1\)
\(R_2\)
\(R_3\)
\(F_h\)
\(F_v\)
\(F_{NW}\)
\(F_{NE}\)
\(\; I\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(R_1\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(R_2\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(R_3\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(F_h\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(F_v\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(F_{NW}\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(F_{NE}\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
Figure4.2.12.Symmetries of a Square Calculator
So does \(D\) with \(\circ\) form a group? You just showed it is closed. We will tell you it is associative. Does \(D\) have an identity? Yes, \(I\text{.}\) Does \(D\) have inverses?
Question4.2.13.
What are the inverses for elements of \(D\text{?}\)
