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Grade 9/10 Math Circles
An Introduction to Group Theory Part 3 - Solutions

Exercise Solutions

Exercise 1

Identify which of the below diagrams are \(4\)-braids.

A description of the diagram follows.

Exercise 1 Solution

Diagrams 1,2,3, and 4 are all \(4\)-braids. At a quick glance, it looks like diagram 4 has a knot, however, if you pull the loop part of the second string towards the right, we get that:

In the original image of the fourth braid, a string starts at left dot 2, goes under the string connecting dot 1 to dot 1, then turns around and goes over the same string again, then crosses over itself to get to right dot 2. In a new image of the same braid, this string starts at left dot 2, goes under the string connecting dot 1 to dot 1, then turns around to go over the same string again but does not cross over itself to arrive at right dot 2.

So, diagram 4 is indeed a \(4\)-braid. Diagram 5 is not a \(4\)-braid because it has a knot (in fact, two knots). Diagram 6 is not a \(4\)-braid because not every dot has a string attached to it.

Exercise 2

Choose a small \(n\), say \(1\leq n\leq 6\), and make \(n\)-braids. Feel free to use a few different \(n\) values.

Exercise 2 Solution

Here are some examples of braids that look a bit more complicated:

Four examples of 6-braids.

Exercise 3

Consider the six \(3\)-braids below. How many different \(3\)-braids are there?

A description of the diagram follows.

Exercise 3 Solution

There are 3 different \(3\)-braids. Diagram 3 becomes diagram 1 by pulling the top string upwards. Diagram 4 becomes diagram 1 by pulling the top and bottom string upwards and downwards, respectively. Similarly, diagram 6 becomes diagram 5 by pulling the top string upwards. Since diagram 5 and 6 have one cross, they are not the same as diagram 1. Diagram 2 is not the same as any other diagram as it has two crosses. Given this, we can take diagram 1,2, and 5 to be the 3 different \(3\)-braids.

Exercise 4

Try to define a binary operation \(\bullet\) on \(B_n\) so that \((B_n,\bullet)\) is a group. Here are some suggestions and comments to help you get started:

Exercise 4 Solution

See Definition 3 from Lesson 3 for an explanation of the binary operation on \(B_n\).

Exercise 5

Convince yourself that \((B_3,\ast)\) is a group.

Exercise 5 Solution

We need to convince ourselves that the 3 group axioms hold for \(B_3\) with concatenation. Let’s go through each axiom:

Axiom 1: For all \(b_1,b_2,b_3\in B_3\), we need \((b_1\ast b_2)\ast b_3\) to be the same \(3\)-braid as \(b_1\ast(b_2\ast b_3)\). This is a bit tough to argue rigorously without introducing a lot of extra notation, so let’s do a concrete example to see how it works. Consider the following \(3\)-braids:

In braid b subscript 1, a string connects left dot 1 to right dot 2, a string that goes under the previous string connects left dot 2 to right dot 1, and a string connects left dot 3 to right fot 3.   In braid b subscript 2, a string connects left dot 1 to right dot 1, a string that goes over then under the previous string connects left dot 2 to right dot 2, and a string connects left dot 3 to right dot 3.    In braid b subscript 3, a string connects left dot 1 to right dot 1, a string connects left dot 2 to right dot 3, and a string that goes under the previous string connects left dot 3 to right dot 2.

We compute \(b_1\ast b_2\) to be

A 3-braid with a string connecting left dot 1 to right dot 2, a string going over the previous string and connecting left dot 2 to right dot 1, and a string connecting left dot 3 to right dot 3.

Concatenating this with \(b_3\), we find that \((b_1\ast b_2)\ast b_3\) is

A 3-braid with a first string connecting left dot 1 to right dot 3, a second string going over the first string and connecting left dot 2 to right dot 1, and a third string going under the first string and connecting left dot 3 to right dot 2.

Similarly, we find that \(b_2\ast b_3\) is

A 3-braid with a first string connecting left dot 1 to right dot 1, a second string going over then under the first string to connect left dot 2 to right dot 3, and a third string going under the second string to connect left dot 3 to right dot 2.

and hence \(b_1\ast(b_2\ast b_3)\) is

A 3-braid with a first string connecting left dot 1 to right dot 3, a second string going over the first string connecting left dot 2 to right dot 1, and a third string going under the first string connecting left dot 3 to right dot 2.

We see that \((b_1\ast b_2)\ast b_3 = b_1\ast (b_2 \ast b_3)\), as desired.

Axiom 2: The identity element in \(B_3\) is the \(3\)-braid

Three horizontal strings connect dots 1,2,3 on the left to corresponding dots 1,2,3 on the right.

If we concatenate this braid with any other \(3\)-braid, say \(b\in B_3\), then the strings of \(b\) just get longer. Since the length of strings does not matter, the result is just \(b\). So, this braid is the identity element in \(B_3\) and we call it \(\text{id}_{B_3}\).

Axiom 3: Lastly, we want to show that each braid in \(B_3\) has an inverse. There is a systematic way to obtain the inverse of any braid in \(B_3\), which we illustrate through concrete examples. Basically, given a braid in \(B_3\), we obtain it’s inverse by flipping the braid over horizontally. To see this, consider the following photo:

Two 3-braids with a vertical line between them. The left braid has left 1 connected to right 2, left 1 connected to right 2 with a string going under the previous one, and left 3 connected to right 3. The right braid has the same connections as the left, but the string connecting left 2 to right 1 goes over instead.

Two 3-braids with a vertical line between them. The left braid has left 1 connected to right 1, left 2 connected to right 2 with a string going over then under the previous one, and left 3 connected to right 3. The right braid has the same connections as the left, but the string connecting left 2 to right 2 goes under then over instead.

Two 3-braids with a vertical line between them. The left braid has left 1 connected to right 1, left 2 connected to right 3, and left 3 connected to right 2 with a string going under the previous one. The right braid has the same connections as the left, but the string connecting left 3 to right 2 goes over instead.

We explain how to obtain the inverses of the \(3\) braids to the left of the red vertical lines. Flipping each of these braids over horizontally is the same as reflecting them in the red vertical line next to them. The result of these reflections are the braids to the right of the vertical lines. The braids on the right are the inverses of the respective braids on the left. We can do this with any braid to obtain it’s inverse!

Problem Set Solutions

  1. Identify which of the below diagrams are \(3\)-braids.

    A description of the diagram follows.

    Solution: Diagrams 1,2,5, and 6 are all \(3\)-braids. Diagram 3 is not a \(3\)-braid because it has a knot. Diagram 4 is not a \(3\)-braid because not every dot has a string attached to it.

  2. Compute the following two concatenations:

    Two 4-braids with an asterisk between them. In both braids, left dots 1,2,3,4 connect to corresponding right dots 1,2,3,4. In the left braid, the second string starts left 2, goes under then over the third string, ending at right 2. In the right braid, the first string starts at left 1, goes over then under the second string, ending at right 1; also, the fourth string starts at left 4, goes under then over the third string, ending at right 4.

    Two identical braids with an asterisk between them. In the braids, left dot 1 connects to right dot 2 and left dot 2 connects to right dot 1 with a string going over the previous one; also, left dot 3 connects to right dot 4 and left dot 4 connects to right dot 3 with a string going under the previous one.

    Solution: The first concatenation is as follows.

    Left dots 1,2,3,4 connect to corresponding right dots 1,2,3,4. The third is horizontal. The second goes under then over the third near its left end. The fourth goes under then over the third near its right end. The first goes over then under the second near its right end.

    The second concatenation is as follows.

    Left dots 1,2,3,4 connect to corresponding right dots 1,2,3,4. The second string goes over then under the first. The fourth string goes under then over the third.

  3. Compute the inverses of the following two braids:

    A 4-braid with left dots 1,2,3,4 connected to corresponding right dots 1,2,3,4. The second string starts at left 2, goes under the third string connecting 3 to 3 and then over the same string, ending at right 2.   A 4-braid with left dots 1,2,3,4 connected to corresponding right dots 1,2,3,4. The first string starts at left 1, goes over the second string connecting 2 to 2 and then under the same string, ending at right 1. The fourth string starts at left 4, goes under the third string connecting 3 to 3 and then over the same string, ending at right 4.

    Solution: The inverse of the first braid is as follows.

    A 4-braid that matches the first braid except the second string goes over then under the third instead of under then over.

    The inverse of the second braid is as follows.

    A 4-braid that matches the second braid except the first string goes under then over the second string, and the fourth string goes over then under the third string.

  4. Consider the set \(B_n\) of all \(n\)-braids, where \(n\in\{1,2,3,\dots\}\). Is \(B_n\) finite or infinite?

    Solution: The set \(B_n\) is finite when \(n=1\) and infinite when \(n\geq 2\). Let’s see why this is true. First consider the case when \(n=1\). There is only one \(1\)-braid, and it is

    One dot on the left connected by a string to one dot on the right.

    This means that there is only one element in \(B_1\), and so \(B_1\) is finite. Next, consider the case when \(n=2\). To see why \(B_2\) is infinite, consider the following \(2\)-braids:

    Six 2-braids, each with a horizontal string connecting left dot 1 to right dot 1. In the first braid, a horizontal line connects left 2 to right 2. In the second, the string starts at left 2, goes over then under the other string, and ends at right 2. In the third, the string starts at left 2, goes over then under, then goes over then under a second time, and ends at 2. This pattern continues with the second string going over and under three times, then four times, etc.

    In all of these \(2\)-braids, we see that the second string wraps around the first string a various number of times. From left to right and top to bottom, we see that the number of wraps increases. From left to right, the \(2\)-braids in the top row have 0 wraps, 1 wrap, and 2 wraps. From left to right, the \(2\)-braid in the bottom row have 3 wraps, 4 wraps, and 5 wraps. We can keep increasing the number of these wraps forever, and so there are an infinite number of \(2\)-braids.
    Lastly, consider the case when \(n\geq 3\). We will use \(B_2\) to show that there are an infinite number of \(n\)-braids. We construct \(n\)-braids as follows. Take the last \(n-2\) strings and dots to be

    One dot on the left connected by a string to one dot on the right.

    Then, we can think of the the remaining string connections (aka top 2 strings and pairs of dots) as a \(2\)-braid. In other words, we have an "embedding" of \(B_2\) into \(B_n\). By varying the top two string connections through the elements of \(B_2\), we generate an infinite number of \(n\)-braids (since \(B_2\) is infinite). So, \(B_n\) is infinite. Here are some concrete examples of \(3\)-braids constructed in this way:

    Six 3-braids, each horizontal strings connecting dot 1 to dot 1 and dot 3 to dot 3. In the first braid, the second string has 0 wraps. In the second braid, the second string has 1 wrap around the first string. In the third braid, the second string has 2 wraps around the first string. This pattern continues in the next three braids adding more wrap each time.

    We see that the top two string connections are just the \(2\)-braids from above.