Section Week 5 (9/22-9/26)
This week we start investigating groups and rings and their properties in depth.
Handout Monday 9/24
Consider the symmetries of a square. By this we mean a transformation that maps the square to itself. We find that there are a total of eight such symmetries: four rotations and four reflections (or mirrors or flips).
Let \(r_0\text{,}\) \(r_1\text{,}\) \(r_2\text{,}\) \(r_3\) be rotations counter-clockwise by 0, 90, 180, and 270 degrees (respectively).
We will use \(f_1\) for a reflection (flip) over the horizontal axis (through opposite sides of the square), \(f_2\) for a vertical reflection (through the line bisecting the top and bottom edges of the square), \(f_3\) for a flip over the line \(y = x\) (from the bottom left to top right of the square) and \(f_4\) for the flip over the other diagonal (bottom right to top left).
We notice that when you compose two symmetries, you still get a symmetry. For example, if you first perform the \(f_1\) symmetry, and then \(r_1\text{,}\) this gives you the same result as performing the single symmetry \(f_3\text{.}\) We will write this as \(r_1f_1 = f_3\text{.}\) Notice that we think of this as \(r_1(f_1(□))\text{;}\) we perform the transformation on the right first, then the one of the left.
Here is the table for the group of eight symmetries of the square. This is called the Dihedral Group on eight elements, which we will write as \(D_4\text{.}\)
| \(\circ\) | \(r_0\) | \(r_1\) | \(r_2\) | \(r_3\) | \(f_1\) | \(f_2\) | \(f_3\) | \(f_4\) |
|---|---|---|---|---|---|---|---|---|
| \(r_0\) | \(r_0\) | \(r_1\) | \(r_2\) | \(r_3\) | \(f_1\) | \(f_2\) | \(f_3\) | \(f_4\) |
| \(r_1\) | \(r_1\) | \(r_2\) | \(r_3\) | \(r_0\) | \(f_3\) | \(f_4\) | \(f_2\) | \(f_1\) |
| \(r_2\) | \(r_2\) | \(r_3\) | \(r_0\) | \(r_1\) | \(f_2\) | \(f_1\) | \(f_4\) | \(f_3\) |
| \(r_3\) | \(r_3\) | \(r_0\) | \(r_1\) | \(r_2\) | \(f_4\) | \(f_3\) | \(f_1\) | \(f_2\) |
| \(f_1\) | \(f_1\) | \(f_4\) | \(f_2\) | \(f_3\) | \(r_0\) | \(r_2\) | \(r_3\) | \(r_1\) |
| \(f_2\) | \(f_2\) | \(f_3\) | \(f_1\) | \(f_4\) | \(r_2\) | \(r_0\) | \(r_1\) | \(r_3\) |
| \(r_3\) | \(f_3\) | \(f_1\) | \(f_4\) | \(f_2\) | \(r_1\) | \(r_3\) | \(r_0\) | \(r_2\) |
| \(f_4\) | \(f_4\) | \(f_2\) | \(f_3\) | \(f_1\) | \(r_3\) | \(r_2\) | \(r_1\) | \(r_0\) |
Handout Wednesday 9/24
Let’s look at a few more examples of groups. Last time we considered symmetries of a square. What about symmetries of a triangle? Instead of using a physical triangle to investigate this, let’s consider just what happens to the corners and develop some notation for tracking this.
While the symmetries of a triangle correspond to rigid motions of the plane, each corresponds exactly to simply a permutation of the vertices of that triangle. So we can consider the different permutations of three objects, call them \(1\text{,}\) \(2\text{,}\) and \(3\text{.}\)
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Use two-line notation for this.
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It might be easier to use cycle notation though. For example, the permutation that swaps 1 and 2 and leaves 3 fixed can be written as \((12)\text{.}\) The cycle \((132)\) means that 1 goes to position 3, 3 goes to position 2, and 2 goes to position 1.
Handout Friday 9/26
We now know what a group is and have some examples. Much of group theory is about deciding what properties must be true of all examples of groups. In other words, what do the group axioms guarantee must be true.
For example, all the groups we have considered have only one identity element. Every element in a group seems to have only one inverse. These are not axioms of the definition of a group. But must they be true of all groups anyway? We can prove they must be.
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First, prove that the identity element is unique. First maybe think about the Cayley table version.
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Suppose \(e\) and \(e'\) were both identity elements. Then \(ee' = e\) and \(e e' = e'\) so \(e = e'\text{.}\)
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If \(g\) is any element of a group \(G\text{,}\) then the inverse of \(g\) is unique. We denote the inverse \(g\inv\text{.}\)
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Let’s prove this. Say \(x_1\) and \(x_2\) are both inverses of \(g\text{.}\) What can we say about \((x_1 g)x_2\) and about \(x_1(g x_2)\text{?}\) By associativity, they must be the same. But one is \(x_2\) and the other is \(x_1\text{.}\)
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We also noticed that in any row of a Cayley table, no element appears more than once. What would this say in terms of equations?
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Well, look at row \(a\) and suppose some element \(g\) appears twice: in column \(b\) and in column \(c\text{.}\) This would mean that \(ab = ac\text{.}\) Can’t we just cancel the \(a\)’s?
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Yes we can! Multiply both sides by \(a\inv\) on the left.
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Let’s talk about inverses some more. One of the reading questions asked whether \((ab)\inv = a\inv b\inv\text{.}\) It doesn’t! But we do have the \((ab)\inv = b\inv a\inv\text{.}\) What does this even mean?
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This is not saying that you can distribute the exponents, as long as you are careful. This says that the inverse of the element \(ab\) is the element you get as the product of two inverses.
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It is helpful to establish the following lemma: if \(xy = e\) then \(x = y\inv\) and \(y = x\inv\text{.}\) In other words, the only way for the product of two elements to be the identity is if they are inverses of each other.
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To prove this, assume \(xy = e\text{.}\) Now multiply on the right by \(y\inv\text{.}\) Tada. Now start over and multiply on the left by \(x\inv\text{.}\)
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So why is \((ab)\inv = b\inv a\inv\text{?}\) Well, what is \((ab)(b\inv a\inv)\text{?}\) Use associativity.
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We also get that \((a\inv)\inv = a\) using the same trick.
A note about notation: we have been writing things as if the operation were multiplication. This is okay, but we need to remember that we are really using an arbitrary operation. We sometimes write the operation as addition, and then inverses are written as \(-a\text{.}\)
The multiplicative notation is nice because it allows us to use exponents, and get the usual rules for exponents. Mostly.
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Is \((ab)^2 = a^2b^2\text{?}\) Alas no. We would expect there to be problems because \((ab)^2 = abab\text{,}\) but \(a^2b^2 = aabb\text{.}\) We can’t switch the middle \(ba\) to \(ab\) unless the group is abelian.
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In fact, we can show that if \((ab)^2 = a^2b^2\) for all \(a, b \in G\text{,}\) then the group must be abelian. Do this.
