These thoughts came to me when I studied abstract algebra. Some of them are general and some are related to the certain book. I will update this post as I continue studying: so currently it could look crappy and uncompleted. You are welcome to discuss these thoughts in the comments section below. My current level of abstract algebra knowledge is pretty low, so I would be glad to any corrections or suggestions..

"Visual group theory" is the book I used to study abstract algebra from scratch. I used the book edit actual as of November 2023.

Hereinafter I use the following notation and terms:

- Exercise. This refers to the exercises at the end of every book chapter that are provided by the book author. Also I list some "Answers to the book exercises" in the consequent sections that are not given by the book author.

- Question. The question that came to my mind while studying. In the "Questions" section I include thoughts that came to my mind and that are not listed in the book. I didn't get answers and solutions to all my questions so far.

- Section. This refers to the consequent numbered section of the consequent chapter. E. g. Section 1.2 is in the chapter 1.

- The book author. He is Nathan Carter. I will refer to him hereinafter as "the book author".

Chapter 1. What is a group?

Questions

*Question 1.1. Can we apply any generator to any achievable Rubik's cube state?*

By "any achievable state" we mean any cube condition that can be derived from the starting point (with same colored edges) via any combination of generators.

*Solution*

We should be able to apply any generator to any state. This is because in Definition 1.9 Rule 1.8 dictates "Rule 1.8. Any sequence of consecutive actions is also an action."

The plain text of the rule seems like ambiguous. However it is verified by exercise 1.3, the answer to it is also given in the book.

*Question 1.2.*

*Out of rules 1.6-1.8, is there any one redundant?*

I. e. can some rule be derived from others? Are all of them necessary? In the section 1.5.3, the book author suggests that all of the rules are necessary. But what is the prove for that?

*Question 1.3.*

*If generators are reversible, are combined actions reversible?*

Generators are simplest non-breakable actions. Combined actions are any actions created by consequently applying any consequence of generators.

Let's assume Rule 1.7 means "every generator is reversible" instead of "every action is reversible". Does it mean that every action is reversible because every generator is reversible?

*Solution*

Yes it does. Let's take some random action G bringing us from the node A0 to An. This action is the result of applying n generators as follows:

generator g0 brings us from A0 to A1,

generator g1 brings us from A1 to A2,

...

generator gn brings us from An-1 to An.

Let's denote reverse elements for generators as h.

Then

h1 is inverse for g1,

h2 is inverse for g2,

...

hn is inverse for gn.

Since every generator is reversible, we can apply h elements. Let's apply hn to An,...,h1 to A1. We can prove by mathematical induction that this gives us A0 in the end. Coming from An to A0 is an inverse action to G.*Question 1.4.*

*Does a group comply with commutative property?*

Does the order of actions performance affect the result?

*Solution*

In general, groups do not comply with commutative property. E. g. see Exercise 1.4(b).

*Question 1.5.*

*Is chess game a group? Why? What are actions in this case?*

So we need to define an object, object states and generators (simplest actions). E. g. in the book Rubik's Cube example is considered in the Chapter 1. So for Rubik's Cube the object is Cube per se. States are combinations of its edges and their color. And actions are rotations of edges in different directions.

So for chess the object is the chess board and its figures (black and white). States are unique combinations of all the figures (not of each figure) on the board cells. So one state is a matrix 64 board cells X 16 figures, it shows cell position of each figure. One action is a move of a certain figure to a certain cell e. g. queen e1->e5.

The book author mentions in the book that "some of the available moves may become unavailable to the player as the game progresses." That means it is not a group because it does not comply with Rule 1.8: "Any sequence of consecutive actions is also an action." See also the solution for Exercise 1.3.

*Question 1.6.*

*What does really Rule 1.8 mean?*

Rule 1.8 says: "Rule 1.8. Any sequence of consecutive actions is also an action". But what does it really mean?

If consequence of actions is an action what is an action per se? Action is not defined formally in Chapter 1. An action is a function. A function has an input and an output. In our case an input is an object state. An output is another state of the same object.

So, we should not have any problems with an input: we can combine any object state with any action. The question is about an output. If we combine actions we should have valid object state returned. See also

*infra*solutions for Exercise 1.3, Exercise 1.5*,*Exercise 2.17.Based on Exercise 1.3, we could rephrase Rule 1.8 to "any action can be applied to any node and it should give valid output".

*Question 1.7. How many states can the object of Rubik's cube have?*

*Question 1.8. What terminology notation can we use for groups like Rubik's cube description?*

*Solution*

This terminology will be useful for us in discussing similar examples further along the book.

Action - a move that changes a state of an object. In case of Rubik's cube it is rotating its edges.

Generator - an action used to generate other actions via combinations with other generators.

Initial state - a state that our game starts from. The same as starting state, initial node, starting node.

Node - the same as state.

Object - an object that changes its states. In case of Rubik's cube it is the cube itself.

State - a state of an object. In case of Rubik's cube it is a combination of its edges that could be achieved by combination of actions (rotations) starting from the initial state.

*Question 1.9. What is the strict definition of a generator?*

In Section 1.4, the book author says: "...Rule 1.5 gives us actions that generate all the others, and are therefore called generators..." So how can we define a generator term strictly?

Say like that: a generator is an action that cannot be generated by combination of other actions other than itself. Unfortunately this does not work. Take a look at Exercise 1.14 (d). Generators are adding 1 and -1. Adding 5 and adding -6 are actions generated by generators. Adding consequently 5 and -6 to zero gives us -1 which is a generator.

So how can we define a generator term strictly?

*Question 1.10. What does Rule 1.5 really mean?*

In Section 1.2 regarding Observation 1.1, the book author says the following. Chess game does not comply with Observation 1.1 (and hence with Rule 1.5). But it depends on how we look at it; it depends on the point of view. If we say every possible combination of figures on a chess board is a state then we can define a large enough quantity of moves to make it in compliance with Rule 1.5. BTW see Question 1.5

*supra*and Exercise 2.14 also.*Question 1.11. How can we automate searching for Rubik's cube game solution?*

Say we have unlimited calculation capacity on our computer. In a simplest way, we could iterate via all the possible states (combinations) of the game. However it will take a lot of computer resources and time. The book author mentioned in Section 2.1, "To store only the data (from which to later reconstruct text and images) would require approximately 1021 bytes, or one billion terabytes". Is there a faster way? E. g. using some tricks from the group theory?

Answers to the book's exercises

*Exercise 1.1*

An answer to this exercise is also given in the end of the original book.

Finally yes it is a group. Compliance with the group rules is as follows:

Rule 1.5. There is a predefined list of actions that never changes. We have just one action which is swapping coins.

Rule 1.6. Every action is reversible. To reverse swapping coins we just need to swap them again.

Rule 1.7. Every action is deterministic. Swapping coins gives only 2 deterministic results.

Rule 1.8. Any sequence of consecutive actions is also an action. We can swap coins any number of times and get only 2 possible results. More accurately, it seems like the book author means that this rule implies that any consequence of actions is applicable and gives certain result other than nothing. See also Exercise 1.3.

*Exercise 1.2*

Yes, it is a group. The logic is the same as for Exercise 1.3.

*Exercise 1.3*

An answer to this exercise is also given in the end of the original book.

As the book author says in his answer, indeed it is not a group. I agree. However we have a note about the book author's argument. The book author argues, we are not in compliance with Rule 1.8 (any sequence of consecutive actions is also an action).

Indeed, we cannot move marbles from our left pocket to our right pocket twenty times in a row. We have just five marbles in our left pocket. So we can move in a row only five marbles to our right pocket. We cannot move the sixth marble to the right pocket.

Let's name it in other way. We can move sixth marble to our right pocket but the result will be called empty set. Nobody prohibit us to have an empty set as an object state.

But we will not be able to comply with Rule 1.6. It is not clear how to get back from the empty set.

So finally the answer to Exercise 1.3 could be as follows: it is not a group because it doesn't comply with Rule 1.6.

*Exercise 1.4*

(a) 3*2*1 = 6.

(b) Yes, these 2 actions can generate all the possible combinations. That is because we can place any picture to any slot.

(c) Yes it is a group, all the rules 1.5-1.8 are satisfied.

(d) This is similar to Exercise 1.3. It is not a group. We cannot commit the new action twice in a row.

On the other side, the devil is in the details. The exercise uses words like "art", "all art", "piece of art". Say we have nothing hanging on the wall, that means we have no art there, an empty set.. Empty set is often included in other sets. Does "all art" term include "no art"? If yes we have a valid group here.

*Exercise 1.5*

An answer to this exercise is also given in the end of the original book.

After reading the book author's answer, we need to define what is an action

*per se*. If we know what action is we then can count how many action does a group have.So an action is a function that has an input and an output. An input is an object state and an action. An output is new object state.

So the question is as follows. If two different actions give similar result does it mean that we count these actions as the same?

In Chapter 4, the book author removes states from the discussion and replaces them with actions. Indeed, every state is is a result of a certain action. Thus, every state can be named with a name of a consequent action.

Hence, quantity of actions equals to quantity of all the possible object states. I will use this conclusion in the Exercise 1.6

*infra*.

*Exercise 1.6*

Hereby, I use the conclusion from the previous Exercise 1.5

*supra*: quantity of actions equals to quantity of unique object states.

Reference to Exercise 1.1

There are only two possible states hence there are two actions.

Reference to Exercise 1.2

The answer is two, the same as for Exercise 1.1.

Reference to Exercise 1.3

This is not a group. Otherwise we would have six states possible and hence six actions.

Reference to Exercise 1.4

For (a), (b) and (c), the answer is the same as for Exercise 1.4(a). The quantity of actions is 6.

For (d), I will not make the solution difficult and thus I assume it is not a group. So we don't need to calculate anything here.

*Exercise 1.7*

An answer to this exercise is also given in the end of the original book.

(a) The action is do nothing. It doesn't change the object state.

(b) Yes, every group will contain the "do nothing" action. This is because of Rule 1.6. Every action A can be reversed by another action B. Consequence A+B is "do nothing" action.

*Exercise 1.8*

(a) The group from Exercise 1.4 satisfies this requirement. Rubik's cube satisfies it as well.

(b) Exercises 1.1 and 1.2 satisfy this requirement.

(c) Like in Exercise 1.4 say we have three walls. However unlike Exercise 1.4 we have only one piece of art. In the beginning, it hangs on the left wall. We have only one generator action: move the piece of art to the next wall to the right. If the piece of art is already on the most right wall then this action moves it to the left most wall.

(d) Like in (c) above we have everything the same except for that we have four walls instead of three.

(e) We ride a starship. It can move 1000 km straight forward or 1000 km straight backward. Another example is as follows. We have a variable and initially it equals to zero. We can add 1 or substract one from this variable.

*Exercise 1.9*

In Exercise 1.5, we discussed that quantity of actions equals to quantity of all the possible object states.

So to create a group with a certain number of actions we need to architect an object with a consequent number of states. The example is based on Exercise 1.4. Say we need to create a group with x actions.

Then like in Exercise 1.4 say we have x walls (kind of polygon style room). However unlike Exercise 1.4 we have only one piece of art. In the beginning, it hangs on the most left wall. We have only one generator action: move the piece of art to the next wall to the right. If the piece of art is already on the most right wall then this action moves it to the left most wall.

*Exercise 1.10*

An answer to this exercise is also given in the end of the original book.

Let's take the starship from Exercise 1.8(e). Now it is able to move only forward not backwards.

*Exercise 1.11*

Dice could be an example.

*Exercise 1.12*

An answer to this exercise is also given in the end of the original book.

Exercise 1.3 is an example. Another example is as follows. Let's take a starship from Exercise 1.8(e). Now it is prohibited to move far than Milky Way.

*Exercise 1.13*

It is about whole numbers, however the exercise says: "...we might name them things like "add 1" and "add -17...". However "-17" seems like not a whole number. We do not consider whole numbers as containing negative numbers. So if we assume that the exercise means whole numbers, than it is not a group. We cannot satisfy the requirement of reversibility (Rule 1.6).

If the exercise is about integers than it is a group and we are in compliance with all the rules 1.5-1.8. Than the smallest set of generators is {add -1; add 1}.

Let's also clarify the situation by applying the framework of notations from Question 1.8

*supra*.The object in this case is a number placeholder. It has a number value in every current moment.

The state is the number value of the placeholder in the current moment.

Generators are changing the current state by +1 or -1. So there are two generators overall. All the other actions are generated by combining these generators.

Initial state could be any number. Say it is zero.

The question is what does the book author mean by saying "to the one you chose.". If we could pick any number in the current moment regardless of the previous state (previous number) than this creates a lot of mess. So we deem for the purpose of the current solution that we pick the current (not any) number of the current state and apply to it one of two generators. So on the next iteration we also have univocal (not any, not random) certainly defined number etc.

Another question is what does the book author mean by saying "adding any whole number to the one you chose". If "any" means random than it is not a group. Because Rule 1.7 is violated.

*Exercise 1.14*

Like in Exercise 1.13, we need to decide whether whole numbers contain negative numbers or not. Let's assume they do for the purpose of this exercise.

(a) Yes it is a group. All the criteria of Rules 1.5-1.8 are satisfied.

(b) No it is not a group. Rule 1.6 is violated.

(c) No it is not a group. Rule 1.6 is violated. Another note is about the book author says "I do not ask you to consider those actions as generators, but as the complete set of action". So it is not clear how does this statement affect the final solution.

(d) Yes it is a group. All the criteria of Rules 1.5-1.8 are satisfied.

Chapter 2. What do groups look like

Questions

*Question 2.1. How to present Cayley diagram as a table (matrix)?*

Graphical Cayley diagram is good for visualization. However a table is a good tool of exploration and automation of calculations.

*Solution*

Rows names set (x-axis) will be a set of object states which is a set of all the nodes in Cayley diagram. Columns names set (y-axis) is generators (simplest non-breakable actions).

Let's say we have m generators g1, g2, …, gm. For example as the book author mentions in Section 1.2, Rubik's cube will have 6 generators as we can rotate each of six edges clockwise and it is enough to define all the other actions. Also we have n possible states of the object (s1, s2, ..., sn). In case of Rubik's cube, n is enormous, it should be something around zillion. Then our table will be as follows:

*Question 2.2. Can we get from any point to any point in Cayley diagram of a group?*

We are given a group and its Cayley diagram. Let's choose its random node (point). Can we make a path via generators actions to any other node? Can there be a node belonging to a group that is not accessible from our chosen node?

*Solution*

It is assumed that we can get to any target node from so called starting node by applying generators actions. Otherwise we wouldn't include that target node in a group.

E. g. in the Chapter 1 starting node was Rubik's cube with each edge having same color.

According to Definition 1.9, every action is reversible. Hence, we can get from any node to the starting node. In turn

*supra*we noted that we can get from the starting node to any node.*Question 2.3. If we could get from any node to any other node does it mean that every action is reversible?*

Say we have a group defined as follows:

Rule 1.5. There is a predefined list of actions that never changes.

Rule extra. We can get from any state (node) to any other node.

Rule 1.7. Every action is deterministic.

Rule 1.8. Any sequence of consecutive actions is also an action.

This is the Definition 1.9 with one rule replaced. We removed Rule 1.6 "Every action is reversible" and we placed Rule extra instead "We can get from any state (node) to any other node". By "node" we understand here any state which is the result of any combination of generators applied to starting point.

Does it mean that Rule 1.6 should follow from these?

*Solution*

Yes it follows. Let's take 2 random nodes: A and B. Rule extra means that we can always get from A to B and from B to A. Hence, A<->B action is reversible.

*Question 2.4. What if we choose another node for start?*

Let's change the initial element, so we start from any other node of the Cayley diagram. The diagram per se remains the same.

Will it still be a group? Will we still comply with all the group rules?

*Question 2.5 What algorithm can allow us to search for all the possible object states in a given group?*

Say we have the following. We have an object and its initial state. Also we have a set of generators. For any given state and generator pair, we know the result, i. e. we know what new state we get by applying given generator to given state. Sow how can we get a set of all possible states?

*Solution*

E. g. in Section 2.4, the book author provided the following algorithm for the rectangle puzzle: "We branched out from the starting position using each generator, and then branched out from each of those positions using each generator again. If the puzzle had been more complicated, we could have continued this process further, exploring farther and wider until we had found every location in the realm." Let's generalize it for any group. Also let's use table format which is machine friendly, especially for processing large amount of data in case of complex groups like Rubik's cube.

*Supra*in Question 2.1 we noted that Cayley diagram could be presented as a matrix: states x generators. Let's use this format for finding all the possible states.

Say our initial state is s1. And we have m generators g1, g2, …, gm. Let's apply each of g to s1, then we will get n states, n≤m. Then let's apply each generator to each of states s1, s2, ..., sn, so we get a matrix n by m dimensions.

The matrix body will contain some new states in turn. Some of resulting states will belong to the set {s1, s2, ..., sn} and some will not. Those that do not are {sn+1, sn+2, ..., sn+k}. Let's join sets {s1, s2, ..., sn} and {sn+1, sn+2, ..., sn+k}. Then we get a new set {s1, s2, ..., sn+k}. This set will form a new axis for the next matrix.

However one question is still remaining. How do we prove that this algorithm covers absolutely all the states?

*Question 2.6. How can we automatically convert Cayley diagram to a matrix from Question 2.1 and back?*

What universal algorithm can we use?

*Question 2.7. Should Cayley diagram contain generators only or all the possible actions and why?*

From the book it follows that the diagram depicts generators only. Can there be a case when depicting all the actions could be useful also?

Answers to the book's exercises

*Exercise 2.1*

An answer to this exercise is also given in the end of the original book.

We have 2 generators here: horizontal flip and vertical flip. There are two unique actions we can get by combining these generators.

First one is "do nothing". We get it by doing two horizontal flips in a row or two vertical flips in a row. It returns us to the initial state.

The second one is performing two different flips in a row: horizontal and vertical or vertical and horizontal.

All the other actions merely repeat previous mentioned ones.

*Exercise 2.2*

This is similar to Exercise 2.1.

We have 2 generators here: Flipping the first switch and flipping the second switch. There are two unique actions we can get by combining these generators.

First one is "do nothing". We get it by doing two first switch flips in a row or two second switch flips in a row. It returns us to the initial state.

The second one is performing two different flips in a row: first switch and second switch or second switch and first switch.

All the other actions merely repeat previous mentioned ones.

*Exercise 2.3*

An answer to this exercise is also given in the end of the original book.

Yes it can. This is "do nothing" action. Here is the group for example.

The object is a light switch. The only generator is flipping it twice.

This generator presents a Cayley diagram arrow pointing the node to itself.

*Exercise 2.4*

An answer to this exercise is also given in the end of the original book.

The diagram will contain 2 nodes since we have only two possible states. These two nodes will be connected by only one line. This line illustrates initial generator and its reverse action which brings the object to the initial state.

*Exercise 2.5*

With a reference to Exercise 1.4, we consider only (b) clause of it because only it describes a group.

Let's use the recursive method provided by the book author in Section 2.4: Beginning at any one configuration or situation, explore carefully using each generator, one at a time. Explore thoroughly and carefully, making a map as you go and labeling the transitions between states with the moves that cause them. Continue until your map contains no unanswered questions, as described above.

Let's mark three pictures consequently as 1, 2 and 3. Then the room with walls will be denoted as a three symbols string. First symbol is the picture hanging on the left wall, second symbol is the picture hanging on the middle wall and third symbol is the picture hanging on the right wall. E. g. in the "123" string we have the "1" picture on the left wall, "2" picture on the middle wal and "3" on the right wall.

We denote generators on the diagram as follows. Red arrow is swapping the art on the left wall with the art on the center wall, Blue arrow is swapping the art on the center wall with the art on the right wall. All the arrows have 2 pointers on both sides because they are reversible by executing the same generator again.

*Exercise 2.6*

Let's deem Exercise 1.13 is about integers not whole numbers. Than Cayley diagram can be presented as follows. We draw an infinite straight (or not straight) line with all the integers marked on it. Each integer will have two outbound and two inbound arrows for the consequent generators. Following Question 2.6

*supra,*we draw generators only in our diagram. So we don't depict infinitely many arrows for all the possible actions.In the solution for Exercise 1.13 supra we found that there are only two generators: adding 1 and subtracting 1 from the current number. Let's deem adding 1 is a red arrow and subtracting 1 is a blue arrow. Then the diagram will be depicted as follows.

*Exercise 2.7*

Say we have a starting node as whole number x. Then we have two possible states: x and -x. Multiplying by one doesn't change the state; and multiplying by minus one changes the state.

Let's denote multiplying by -1 as red arrow. Multiplying by one doesn't change a state so we don't have an arrow notation for it.

So x is not zero. If it would be zero then we have just one state which is starting node, the zero itself. Otherwise, the diagram is as follows.

*Exercise 2.8*

An answer to this exercise is also given in the end of the original book. Here is the note about the solution provided by the book author. In the solution picture for Exercise 2.8, there are arrows of only two colors. However the exercise provides three generators. That means, the book author uses only two of them. This is more elegant solution. In the book author's solution picture, red arrow stands for quarter turn and blue arrow stands for one of flips (either horizontal or vertical).

(a)

We have 8 possible states here. One way to calculate their number is as follows. Let's pick one number from one of the corners of the square rectangle. Say it is one. This number can take one of four corners of our square. We can do that by merely rotating the figure. So we already have four possible combinations.

In each of these combinations we can have two possible outcomes. We can switch between them by consequently doing quarter turn and horizontal flip. E. g. when number one is in the top left corner we could have the possible two outcomes:

So in total we have 4*2=8 possible nodes in this group

Another way to find the total quantity of combinations is as follows. First we turn it clockwise four times so then we have four combinations. Then we flip the square paper so it lays down on the table on its another side. Then we turn the square clockwise again four times. And here are another four states. Total is eight.

And we have three generators.

Let's denote the diagram generators as following arrows:

Red arrow: flip vertically. This arrow has pointers on the both sides because this flip is reversible by applying the same flip again.

Blue arrow: flip horizontally. This arrow also has pointers on the both sides because this flip is reversible by applying the same flip again

Black arrow: quarter-turn clockwise. This arrow has a pointer only on one side.

So the diagram is as follows:

(b)

In the rectangle puzzle, the quarter turn changes the shape of space occupied by the figure. However it is not clear why we cannot use turn for rectangle puzzle. In fact we can. Nothing prohibits us to change the final space shape occupied by the figure yet. Especially if we rotate rectangle around the geometric center of the figure. Later in Chapter 3 the book author will impose the restriction on a shape occupied by a figure. However for now we are free to rotate a rectangle as well as a square.

*Exercise 2.9*

An answer to this exercise is also given in the end of the original book.

On a Cayley diagram, let's denote arrow colors as follows:

Vertical flip - red arrow.

Horizontal flip - blue arrow.

Both flips - black arrow.

(a)

(b)

Hereby, the diagram shows that v and b are sufficient to generate V4 because of the following: we can get from any point to any other point on the diagram with the current arrows only. By the way, see also Question 2.3

*supra*.(c)

Hereby, like in (b)

*supra*the diagram shows that h and b are sufficient to generate V4 because of the following: we can get from any point to any other point on the diagram with the current arrows only. See also Question 2.3 supra.

*Exercise 2.10*

We have a clock hanging on the wall. The only generator is moving clock arrows 12 hours forward. This generator is reversible by itself. By doing this generator two times, we come back to the initial state because we move arrows 24 hours forward.

*Exercise 2.11*

Before looking for possible examples, let's find out what are common properties of the rectangle Cayley diagram from Figure 2.9. So it would be easier to look for a practice situation if we know its criteria (properties).

We see from the diagram that:

1) There are two generators.

2) If we apply any generator to any node two times we get "do nothing" action.

So we need to model a situation where we have two generators which are reversible by themselves.

Our planet rotates around Sun. At the same time, it rotates around itself (around its own axis). Let first generator be to rotate the planet 180 degree around the sun (make half circle). So the second generator is to rotate the planet 180 degree around its own axis (half circle also). Applying any of these generators two times gives us 360 circle which means "do nothing" action.

*Exercise 2.12*

An answer to this exercise is also given in the end of the original book.

Like in the solution to Exercise 2.10 supra, we have a clock hanging on the wall. The only generator is moving clock arrows 8 hours forward. So by doing this generator three times, we move arrows 24 hours forward. So we are in the initial state again.

*Exercise 2.13*

An answer to this exercise is also given in the end of the original book.

Generators are arrows on a Cayley diagram. It seems like we don't show actions on Cayley diagram that are combined of generators. So we show there generators only. See also Question 2.6

*supra*.

*Exercise 2.14*

There is a predefined list of arrows (generators) on a Cayley diagram that never changes. For the second part of the exercise we cannot give a clear answer yet. It is not 100 percent clear what does Rule 1.5 really mean, see for details Question 1.10

*infra*.

*Exercise 2.15*

Say we can get from A node to B node via some arrow (generator) on the Cayley diagram. This path should be reversible. That means, we should be able to get from B node back to A node via certain path of arrows.

Here is the simplest possible Cayley diagram not satisfying Rule 1.6:

An answer to this exercise is also given in the end of the original book.

*Exercise 2.16*

Every arrow is univocal, it is determined and it points from one node to only one another node, regardless of random external circumstances. So the arrow behavior is determined and it doesn't depend on external random circumstances.

Let's draw an example not complying with Rule 1.7. Same generator points simultaneously from A node to two other nodes. Every time the choice between two nodes is made on the external random circumstances. Say it is based on the random numbers generator or a side of dropped coin.

*Exercise 2.17*

An answer to this exercise is also given in the end of the original book.

See also Question 1.6

*supra*. Every node should have outbound arrows for all the generators. Otherwise, this Cayley diagram is not a group. E. g. like in Exercise 2.15*supra,*this diagram is not in compliance with Rule 1.8:

*Exercise 2.18*

Before coming to the solution, let's define its criteria like in Exercise 2.11

*supra*. The previous square rectangle game from Exercise 2.8 has the following properties:(0) Compulsory requirement: generators should form a group as per Rules 1.5-1.8.

(1) Compulsory requirement: Any action should not change the shape of the initial placeholder. After moving, the rectangle should fit the initial placeholder shape again.

(2) If possible, quantity of generators should be minimized. Like the book author did it in his solution for Exercise 2.8. Though there is no official requirement for minimization.

(2) If possible, quantity of generators should be minimized. Like the book author did it in his solution for Exercise 2.8. Though there is no official requirement for minimization.

(3) If possible, generators should be defined in the following way. They should cover all the possible combinations of the triangle figure complying with (1) clause

*supra*. Note we can form a group not complying with this optional requirement: say we have a rectangle puzzle from Section 2.2 with only one action of horizontal flip.

All medians of equilateral triangle intersect in the same point called a centroid. We need medians and centroid to define generators for this group. The following possible actions comply with (1):

*-*Turn the triangle clockwise 120 degree around its centroid.

- Flip the triangle around its any median.

So with just two generators we can generate a group of equilateral triangle. These two could be flips around two different medians. Or they could be the flip around one median and rotation 120 degree around the centroid.

For the map of the group please refer to the book author's diagrams

*supra*in Exercise 2.19(c).

*Exercise 2.19*

(a)

In both a triangle and a square it was enough to have two generators. So the conjecture is two.

(b)

Let's build bisectors from all of the pentagon corners. Their intersection is the centroid of the pentagon. Like in Exercise 2.18 supra, we take two actions. First is flipping the pentagon around one it's bisectors. Second is rotating the pentagon clockwise 72 degrees.

So our conjecture is confirmed. These two actions are enough to cover all the possible states of this group.

The open question is as follows. Can we use two generators to generate the pentagon group by choosing flips around two different medians (bisectors)? So we don't use 72 degrees rotation at all. We can try it by cutting a piece of pentagon paper and play with it like we did with rectangle square

*supra*(like the book author proposed in Section 2.2).*Or we can try to prove this conjecture for the common case which seems like more difficult. Anyway this question is still open..*For the map of the group please refer to the book author's diagrams

*supra*in Exercise 2.19(c).(c)

Here are the diagrams from the Group Explorer's library.

The equilateral triangle group:

The square group:

The regular pentagon group:

(ii) All the diagrams do support the conjecture of that two generators are enough to generate a group for each case. There are only arrows of two colors on each diagram.

(d)

In the (b) clause

*supra*we considered two cases for the conjecture:- Generators are flips around two different medians. This should work for a triangle, see Exercise 2.18

*supra*. However we do not provide a proof for this case hereby. There are some difficulties. There is a situation when a polygon has even number of vertices. E. g. look at the square from Exercise 2.8. Flipping around two medians is not enough to cover all the possible positions of the placeholder shape.- Generators are a flip around one median and a rotation for 360/n degrees around a centroid where n is a number of polygon vertices. We will provide a proof for this case.

We provide a proof that our conjecture complies with criteria (0)-(3) defined in the solution to Exercise 2.18

*supra*.(0) We form a group as per Rules 1.5-1.8:

Rule 1.5 is satisfied. There is a predefined list of actions that never changes. Well yep, there is.

Rule 1.6 is satisfied. Every action is reversible. To reverse a flip we need to do a second flip in a row. To reverse a rotation we need to do another n-1 rotations after that. So we are kind of doing full circle and returning to the initial position. Since generators are reversible, any combined actions are also reversible, see also Question 1.3

*supra*.Rule 1.7 is satisfied. Every action is deterministic. Well yep, they are.

Rule 1.8 is satisfied. Any sequence of consecutive actions is also an action. See also Question 1.6

*supra*. Any generator results in the polygon taking the same shape of the placeholder due to the symmetry of the polygon. So we can apply any generator to this shape again.(1) Generators do not change the shape of the initial placeholder. Yep, this is due to the symmetry of the polygon.

(2) Quantity of generators is minimized. Currently the quantity is two. With the quantity of one we can still generate a group as per (0) clause

*supra*. However the question is open as whether we are able to generate all the possible positions of the figure in the placeholder shape as as per clause (3)*infra*. Theoretically we could define a generator as a consequence of IF algorithm statements like for pentagon say: if it is iteration 1 then rotate, if it is iteration 6 then flip etc. However I assume it will violate Rule 1.7 in the book author's meaning. So the question bout compliance with this (2) clause is still open.(3) Generators cover all the possible combinations of the polygon figure complying with (1) clause

*supra.*Number of states for n-gon equals to 2n. We can get them as follows. By rotating the figure n times by 360/n degrees each time, we get first n positions.The open question: is 2n the maximum? We can prove that it is using proof to the contrary. Say we have a valid figure position not belonging to our 2n set. At least one of our n-gon corners should lay outside of the n-gon placeholder. This is because our 2n set covers all the cases where figure's corners lay inside the placeholder. However the situation of the corner laying outside the placeholder is impossible because the n-gon should belong to the placeholder and be equal to it as per the (1) clause supra.

Chapter 3. Why study groups?

Questions

*Question 3.1. What is the real practical use of groups?*

In Chapter 1, we saw that combinations of Rubik's cube can form something called a group. In chapter 3, we saw that the same can draw molecules of a crystal and pattern on a fence. But why do we need it? Can groups help us to solve Rubik's cube puzzle? Can they help us to see some crystal structure that we cannot see in the microscope?

*Question 3.2. Should we improve the step 1 in Definition 3.1 so it becomes stricter and less vague?*

Step 1 asks to identify all equal parts of the shape. Instead, we could say "Find a way to break up the shape into the

*maximum*number of equal parts. Parts are considered equal if (1) their figures are equal and (2) they are surrounded by the equal figures (spaces)". Namely:(1) Figures are equal in classical geometric terms. E.g. two triangles are equal if one side and two adjacent angles of one triangle equal to consequent analogues of another triangle.

(2) Figures are surrounded by equal spaces so one figure can be easily replaced by another together with surrounding spaces. E. g. the square with a side length of 4 consists of 16 equal squares with a side of 1. However we can only break down the square into four equal squares surrounded by equal spaces.

The requirement of

*maximum*when counting the quantity of equal parts is necessary here. E.g. we can break equilateral triangle in two equal parts (laying on both sides of any median) as well as in three equal parts.BTW, the book author also aims to find the maximum parts. See e. g. discussion in Section 3.1.3, Figure 3.9. So in that case we take half leaf not the whole leaf because we need

*maximum*equal parts. Also we can't take quarter leaf because the parts should be surrounded by*equal*spaces.*Question 3.3. Should we improve the step 2 in Definition 3.1 so it becomes stricter and less vague?*

Step 2 asks to identify possible actions that save a placeholder for a shape to be moved. We can define it instead as "identify the

*minimum*list of actions that cover all the possible shape positions within its placeholder". There is only one possible quantity number for the count of the minimum list of actions. However there could be multiple possible lists of actions complying with the given criteria. E. g. for Figure 3.2 (discussed by the author of the book in Section 3.1.1), first option is the action of rotating clockwise and second option is the action of rotating counterclockwise. Both of them give the same results in terms of possible positions sets.*Question 3.4. How do we define symmetry?*

We intuitively understand that human left and right hands and legs look similar. Thus, human is symmetric. But how can we define symmetry on a more strict level as a math term not as a merely intuitive feeling?

*Solution*

To define symmetry, let's define geometric motion first. Roughly we can define it as follows. Motion of the old figure is creating a new figure somewhere in the space and deleting the old figure so that the new figure equals to the new figure. Symmetry is ability to perform motion where the placeholder for the new figure equals to the placeholder for the old figure. So any dot of the new figure belongs to the shape of the old figure and no dot of the new figure is out of the shape of the old figure.

*Question 3.5. For Question 3.2 supra, how can we prove that the current quantity of parts is maximum possible?*

Is there a universal proof? If not, what are for are the proofs for specific examples discussed in the book? E. g. friezes in Section 3.1.3.

*Question 3.6. For Question 3.3 supra, how can we prove that the current quantity of generators is minimum possible?*

Is there a universal proof? If not, what are for are the proofs for specific examples discussed in the book? E. g. friezes in Section 3.1.3.

*Question 3.7. For every infinite pattern, can we create a finite analogue with similar generators?*

So it could be simpler to implement the algorithm from Definition 3.1.

*Question 3.8. In Definition 3.1 step 1 how can we algorithmize the process?*

Is there some methodology that helps us to find maximum parts easily and logically not only intuitively? So we can apply it to any geometric figure?

*Question 3.9. In Definition 3.1 step 2 how can we algorithmize the process?*

Is there some methodology that helps us to find minimum set of actions easily and logically not only intuitively? So we can apply it to any geometric figure?

Not a complete solution yet, just a thought. In step 1 we divided the figure into parts a1,a2,...,an. Take one part a1 (or it could be any ai where 1<i<n). Then iterate through all the remaining parts aj, 1<j<n, j≠i. For every aj, move the figure so ai replaces aj and the figure fills in its initial placeholder.

*Question 3.10.*

*Other than Definition 3.1, is there other way to create a group based on symmetry?*

Say we have a figure that has some symmetry. How can we create a group based on that? Is algorithm from Definition 3.1 the only one?

*Question 3.11. Can the group be generated based on non-symmetric geometric figure?*

So we define motions for a figure which become actions for a group. Can we do that without symmetry and why?

Answers to the book's exercises

*Exercise 3.1*

*Exercise 3.2*

*Exercise 3.3*

*Exercise 3.4*

*Exercise 3.5*

*Exercise 3.6*

An answer to this exercise is also given in the end of the original book.

*Exercise 3.7*

*Exercise 3.8*

An answer to this exercise is also given in the end of the original book.

*Exercise 3.9*

*Exercise 3.10*

*Exercise 3.11*

An answer to this exercise is also given in the end of the original book.

*Exercise 3.12*

*Exercise 3.13*

An answer to this exercise is also given in the end of the original book.

*Exercise 3.14*

*Exercise 3.15*

*Exercise 3.16*

Chapter 4. Algebra at last

Questions

*Question 4.1. Can number of columns in a multiplication table be lower?*

Multiplication tables are described in the Section 4.3 of the Carter book. It is defined in the book, that number of rows in a table is equal to number of columns which is the total number of actions. However to define a table in a unique way we need only generators as columns. So we don't need to include all the actions in columns.

*Question 4.2. How can creation of multiplication tables be automated?*

It is obvious that we can write a customized piece of code for every certain group. But is there a universal algorithm that can be applied to any group? So we don't customize it every time?

If we talk about certain group - can we automate it via formulas in Google Sheets or MS Excel? This is much easier than writing code.

Without automation, a lot of exercises result in monkey job...

See also

*infra*solution to Exercise 4.8 in part of Reference to Exercise 2.8. At least partially we could ease monkey job in that solution. However some is still remaining.

*Question 4.3. Are all 4 of the Definition 4.2 clauses necessary?*

Out of clauses 1-4, is there any one redundant? I. e. can some property be derived from others?

*Question 4.4. Can we take any node as a starting point in the diagram of actions?*

When creating a diagram of actions, we need to choose the initial element. As per the Definition 4.1, we can choose any element. Why any? Will it still be a group as per Definition 1.9 regardless of which element we choose as a starting node?

*Question 4.5. Can 2 row headers of a multiplication table be the same?*

Can the same node appear 2 times in one axis of a multiplication table?

*Solution*

The answer is nope. Proof? Rule 1.7 provides that every action is deterministic. Let's prove that any consequence of actions is deterministic either. To be continued...

*Question 4.6. For the same given group with the same starting node, can there be 2 different multiplication tables?*

E. g. axis nodes (apart from the starting node) could differ, table contents could differ.

*Solution*

Short answer is yes. Order of nodes in the axis could differ. If we denote a node by actions path, sometimes we could come to the same node via 2 different paths. See e. g. the solution to Exercise 4.8

*infra,*references to Exercises 2.5 and 1.4(b). We can come to the 321 node via 2 different ways: lrl and rlr.Answers to the book's exercises

*Exercise 4.1*

*Exercise 4.2*

*Exercise 4.3*

*Exercise 4.4*

*Exercise 4.5*

An answer to this exercise is also given in the end of the original book.

*Exercise 4.6*

*(c)*

*Exercise 4.7*

*Exercise 4.8*

Reference to Exercise 2.4 and 1.1

Reference to Exercises 2.5 and 1.4(b)

Let's take the following notation:

We have 3 pictures denoted with the numbers 1, 2, 3.

We denote states (nodes) as:

- the "123" state is: "1" picture is on the left, "2" picture is in the center, "3" picture is on the right;

- the "132" state is: "1" picture is on the left, "3" picture is in the center, "2" picture is on the right.

- etc.

We denote generators actions as:

- "n" is no action, an identity;

- "l" is swapping a picture on the left with a picture in the center;

- "r" is swapping a picture on the right with a picture in the center.

As it is more convenient, I draw multiplication table in 2 steps. First, I create a table with nodes names like "123". After that, I convert these names to actions like "lrl". Useful function to do that in Google Sheets is vlookup.

Reference to Exercises 2.6 and 1.13

If Exercise 1.13 is about whole numbers than it has no solution, see also the answer to Exercise 1.13

*supra*. If it is about integers than the multiplication matrix is as follows:Reference to Exercise 2.7 and 1.14 part (d)

Reference to Exercise 2.8

Let's make the following notation for the actions generators:

N - no action

R - rotate quarter clockwise

H - flip horizontally

V - flip vertically

Then matching between the rectangle positions and actions is as follows:

Let's automate creation of a multiplication table in Google Sheets (it could be MS Excel or any other spreadsheet software as well). We fill the table as follows. Each cell is joined row header and column header. E. g. in the C4 cell formula is =$A4&C$1. Then we remove formulas and leave values only in the sheet. We will get something like that:

In the section 4.2 "Combine, combine, combine" of the book there is a technique of simplification formulas. We will use the same. Let's find which combinations of actions equal to N (no action), So we have

NN = N

RRRR = N

HH = N

VV = N

RVRV = N

RHRH = N

There are more combinations resulting in N, but we will skip them because we don't meet them in the table

*supra*. So we shorten all the mentioned combinations from the table. We can do that because of associativity property of this group. Why this group is associative? (we don't list a proof here) Also we remove excessive N. We have:Then we need to match table cells with row headers or column headers. So each cell should be equal to some row or column header. Unfortunately, some monkey job is here. So finally we have

Sorry if I did some mistakes while filling it in. Any feedback is greatly appreciated.

*Exercise 4.9*

*Exercise 4.10*

An answer to this exercise is also given in the end of the original book.

*Exercise 4.11*

*Exercise 4.12*

*Exercise 4.13*

An answer to this exercise is also given in the end of the original book.

*Exercise 4.14*

An answer to this exercise is also given in the end of the original book.

*Exercise 4.15*

*Exercise 4.16*

*Exercise 4.17*

An answer to this exercise is also given in the end of the original book.

*Exercise 4.18*

An answer to this exercise is also given in the end of the original book.

*Exercise 4.19*

An answer to this exercise is also given in the end of the original book.

*Exercise 4.20*

*Exercise 4.21*

*Exercise 4.22*

An answer to this exercise is also given in the end of the original book.

*Exercise 4.23*

An answer to this exercise is also given in the end of the original book.

*Exercise 4.24*

An answer to this exercise is also given in the end of the original book.

*Exercise 4.25*

*Exercise 4.26*

An answer to this exercise is also given in the end of the original book.

*Exercise 4.27*

An answer to this exercise is also given in the end of the original book.

*Exercise 4.28*

*Exercise 4.29*

*Exercise 4.30*

*Exercise 4.31*

*Exercise 4.32*

An answer to this exercise is also given in the end of the original book.

*Exercise 4.33*

An answer to this exercise is also given in the end of the original book.

Chapter 5. Five families

Questions

Answers to the book's exercises

*Exercise 5.1*

*Exercise 5.2*

*Exercise 5.3*

*Exercise 5.4*

*Exercise 5.5*

*Exercise 5.6*

An answer to this exercise is also given in the end of the original book.

*Exercise 5.7*

*Exercise 5.8*

*Exercise 5.9*

An answer to this exercise is also given in the end of the original book.

*Exercise 5.10*

*Exercise 5.11*

*Exercise 5.12*

*Exercise 5.13*

*Exercise 5.14*

*Exercise 5.15*

*Exercise 5.16*

*Exercise 5.17*

*Exercise 5.18*

*Exercise 5.19*

*Exercise 5.20*

An answer to this exercise is also given in the end of the original book.

*Exercise 5.21*

*Exercise 5.22*

*Exercise 5.23*

An answer to this exercise is also given in the end of the original book.

*Exercise 5.24*

An answer to this exercise is also given in the end of the original book.

*Exercise 5.25*

*Exercise 5.26*

*Exercise 5.27*

*Exercise 5.28*

An answer to this exercise is also given in the end of the original book.

*Exercise 5.29*

*Exercise 5.30*

An answer to this exercise is also given in the end of the original book.

*Exercise 5.31*

An answer to this exercise is also given in the end of the original book.

*Exercise 5.32*

*Exercise 5.33*

*Exercise 5.34*

*Exercise 5.35*

An answer to this exercise is also given in the end of the original book.

*Exercise 5.36*

*Exercise 5.37*

*Exercise 5.38*

An answer to this exercise is also given in the end of the original book.

*Exercise 5.39*

*Exercise 5.40*

An answer to this exercise is also given in the end of the original book.

*Exercise 5.41*

*Exercise 5.42*

*Exercise 5.43*

An answer to this exercise is also given in the end of the original book.

*Exercise 5.44*

An answer to this exercise is also given in the end of the original book.

Chapter 6. Subgroups

Questions

Answers to the book's exercises

*Exercise 6.1*

An answer to this exercise is also given in the end of the original book.

*Exercise 6.2*

*Exercise 6.3*

An answer to this exercise is also given in the end of the original book.

*Exercise 6.4*

*Exercise 6.5*

*Exercise 6.6*

*Exercise 6.7*

*Exercise 6.8*

*Exercise 6.9*

*Exercise 6.10*

*Exercise 6.11*

*Exercise 6.12*

An answer to this exercise is also given in the end of the original book.

*Exercise 6.13*

*Exercise 6.14*

An answer to this exercise is also given in the end of the original book.

*Exercise 6.15*

*Exercise 6.16*

An answer to this exercise is also given in the end of the original book.

*Exercise 6.17*

An answer to this exercise is also given in the end of the original book.

*Exercise 6.18*

*Exercise 6.19*

*Exercise 6.20*

*Exercise 6.21*

*Exercise 6.22*

*Exercise 6.23*

An answer to this exercise is also given in the end of the original book.

*Exercise 6.24*

*Exercise 6.25*

*Exercise 6.26*

An answer to this exercise is also given in the end of the original book.

*Exercise 6.27*

*Exercise 6.28*

An answer to this exercise is also given in the end of the original book.

*Exercise 6.29*

*Exercise 6.30*

An answer to this exercise is also given in the end of the original book.

*Exercise 6.31*

Chapter 7. Products and quotients

Questions

Answers to the book's exercises

*Exercise 7.1*

*Exercise 7.2*

*Exercise 7.3*

An answer to this exercise is also given in the end of the original book.

*Exercise 7.4*

An answer to this exercise is also given in the end of the original book.

*Exercise 7.5*

*Exercise 7.6*

*Exercise 7.7*

*Exercise 7.8*

An answer to this exercise is also given in the end of the original book.

*Exercise 7.9*

An answer to this exercise is also given in the end of the original book.

*Exercise 7.10*

*Exercise 7.11*

*Exercise 7.12*

*Exercise 7.13*

*Exercise 7.14*

An answer to this exercise is also given in the end of the original book.

*Exercise 7.15*

An answer to this exercise is also given in the end of the original book.

*Exercise 7.16*

An answer to this exercise is also given in the end of the original book.

*Exercise 7.17*

*Exercise 7.18*

An answer to this exercise is also given in the end of the original book.

*Exercise 7.19*

*Exercise 7.20*

*Exercise 7.21*

An answer to this exercise is also given in the end of the original book.

*Exercise 7.22*

*Exercise 7.23*

*Exercise 7.24*

*Exercise 7.25*

An answer to this exercise is also given in the end of the original book.

*Exercise 7.26*

*Exercise 7.27*

*Exercise 7.28*

*Exercise 7.29*

*Exercise 7.30*

*Exercise 7.31*

*Exercise 7.32*

*Exercise 7.33*

An answer to this exercise is also given in the end of the original book.

*Exercise 7.34*

*Exercise 7.35*

*Exercise 7.36*

*Exercise 7.37*

Chapter 8. The power of homomorphisms

Questions

Answers to the book's exercises

*Exercise 8.1*

*Exercise 8.2*

*Exercise 8.3*

*Exercise 8.4*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.5*

*Exercise 8.6*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.7*

*Exercise 8.8*

*Exercise 8.9*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.10*

*Exercise 8.11*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.12*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.13*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.14*

*Exercise 8.15*

*Exercise 8.16*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.17*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.18*

*Exercise 8.19*

*Exercise 8.20*

*Exercise 8.21*

*Exercise 8.22*

*Exercise 8.23*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.24*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.25*

*Exercise 8.26*

*Exercise 8.27*

*Exercise 8.28*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.29*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.30*

*Exercise 8.31*

*Exercise 8.32*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.33*

*Exercise 8.34*

*Exercise 8.35*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.36*

*Exercise 8.37*

*Exercise 8.38*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.39*

*Exercise 8.40*

*Exercise 8.41*

*Exercise 8.42*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.43*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.44*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.45*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.46*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.47*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.48*

*Exercise 8.49*

An answer to this exercise is also given in the end of the original book.

*Exercise 8.50*

An answer to this exercise is also given in the end of the original book.

Chapter 9. Sylow theory

Questions

Answers to the book's exercises

*Exercise 9.1*

*Exercise 9.2*

*Exercise 9.3*

An answer to this exercise is also given in the end of the original book.

*Exercise 9.4*

*Exercise 9.5*

*Exercise 9.6*

*Exercise 9.7*

*Exercise 9.8*

*Exercise 9.9*

*Exercise 9.10*

*Exercise 9.11*

*Exercise 9.12*

An answer to this exercise is also given in the end of the original book.

*Exercise 9.13*

*Exercise 9.14*

*Exercise 9.15*

An answer to this exercise is also given in the end of the original book.

*Exercise 9.16*

An answer to this exercise is also given in the end of the original book.

*Exercise 9.17*

An answer to this exercise is also given in the end of the original book.

*Exercise 9.18*

*Exercise 9.19*

*Exercise 9.20*

*Exercise 9.21*

An answer to this exercise is also given in the end of the original book.

*Exercise 9.22*

An answer to this exercise is also given in the end of the original book.

*Exercise 9.23*

An answer to this exercise is also given in the end of the original book.

*Exercise 9.24*

*Exercise 9.25*

*Exercise 9.26*

*Exercise 9.27*

An answer to this exercise is also given in the end of the original book.

*Exercise 9.28*

An answer to this exercise is also given in the end of the original book.

Chapter 10. Galois theory

Questions

Answers to the book's exercises

*Exercise 10.1*

*Exercise 10.2*

*Exercise 10.3*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.4*

*Exercise 10.5*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.6*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.7*

*Exercise 10.8*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.9*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.10*

*Exercise 10.11*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.12*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.13*

*Exercise 10.14*

*Exercise 10.15*

*Exercise 10.16*

*Exercise 10.17*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.18*

*Exercise 10.19*

*Exercise 10.20*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.21*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.22*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.23*

*Exercise 10.24*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.25*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.26*

*Exercise 10.27*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.28*

An answer to this exercise is also given in the end of the original book.

*Exercise 10.29*

*Exercise 10.30*

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