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This is an exercise in the book "A book of abstract algebra" by Pinter, 3E.

Let's devise a coin game.

Imagine there are two coins, each can be placed in either location $A$ or $B$. Moreover, each coin can be flipped.

Define 8 moves:

$M_1$ - flip over the coin at $A$

$M_2$ - flip over the coin at $B$

$M_3$ - flip over both coins

$M_4$ - switch the coins

$M_5$ - flip the coin at A, then switch

$M_6$ - flip the coin at B, then switch

$M_7$ - flip both coins, then switch

$I$ - do not change anything

Example: $M_4 * M_1 = M_2 * M_4 = M_6$

"switch" means "change places"

I will spare you from what the exercise is asking to do. But I have noticed something playing with the problem. One of the aspects of the definitions of an operation implies that if $*$ is an operation, then $a*b$ is unambiguously and uniquely defined. An operation on the above set $G = \{ M_1, M_2, M_3, M_4, M_5, M_6, M_7, I \}$ has been defined as "performing any two moves in succession". Imagine now that we have done: $M_1 * M_2$, well then the result of this operation can be expressed with two elements in the set, namely: $M_1 * M_2 = M_3 = M_7$ and so this creates ambiguity as to what this specific operation ("performing any two moves in succession") assigns two members of the set ($M_1, M_2$) to. As such, can we even consider $\langle G, *\rangle$ a group (I obviously must be missing / misunderstanding something)?

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    $\begingroup$ $M_1 * M_2 \neq M_7$: $(M_1 * M_2)(H, T) = M_1(H, H) = (T, H) \neq (H, T) = M_7(H, T)$ (writing $H$ for heads and $T$ for tails). $\endgroup$ Commented May 18 at 14:51
  • $\begingroup$ I believe the coins are supposed to be different. Otherwise it doesn't make sense to have $M_4$ as it is the same as $I$. $\endgroup$ Commented May 19 at 14:55
  • $\begingroup$ @RossMillikan A state of the game does not need to include which coin is where - only what the coins at each location are showing. $\endgroup$
    – tkf
    Commented May 19 at 17:29

3 Answers 3

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The game has $4$ states, given by what the coin at $A$ is showing, and what the coin at $B$ is showing. We may arrange these four states as the vertices of a square as follows:

enter image description here

Here the position of a vertex on the horizontal axis is determined by the coin at $A$ and the position of a vertex on the vertical axis is determined by the coin at $B$.

Now $M_1$ is a reflection about the vertical axis through the center of the square. $M_2$ is a reflection about the horizontal axis through the center of the square. $M_4$ is a reflection about the southwest-northeast diagonal.

enter image description here

The other actions are compositions of $M_1, M_2, M_4$. In particular $M_3$ is a $180^\circ$ rotation about the center of the square. On the other hand $M_7$ is a reflection about the northwest-southeast diagonal. Thus $M_3\neq M_7$.

Perhaps your confusion was caused because $M_3$ and $M_7$ agree on two of the states. However they disagree on the other two states, so they are different.

Overall, as a group $\{I,M_1,M_2,M_3,M_4,M_5,M_6,M_7\}$ is just the symmetry group of a square: the dihedral group of order $8$.

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    $\begingroup$ I see, indeed that was my confusion. I see now that if we pick a different initial state such as (Tails, Heads), then $M_3$ produces (Heads, Tails), whereas $M_7$ produces (Tails, Heads). $\endgroup$
    – nz_
    Commented May 19 at 20:57
  • $\begingroup$ Are these plots you have produced a standard way to analyse small-ish groups? I haven't seen this type of analysis in my book, yet. $\endgroup$
    – nz_
    Commented May 19 at 20:58
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    $\begingroup$ and so going to back to the point about coins being indistinguishable, it does seem that that was / is the intention of the question, thank you for clarifying $\endgroup$
    – nz_
    Commented May 19 at 20:59
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    $\begingroup$ It is often useful to find geometric actions of a group, to help understand it, so in that sense yes it is a standard way to analyse groups. This approach leads into a field of mathematics called representation theory. $\endgroup$
    – tkf
    Commented May 20 at 0:38
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Edit: The first sentence in my answer is correct, so in a sense that answers the OP's question. But the rest of the argument isn't right. I'm leaving it as a cautionary tale.

@tkf 's answer explains what's going on.

(I did know this was the dihedral group.)


$M_3$ and $M_7$ are not the same. You can see that with a physical experiment with two different coins each of which has distinguishable sides.

(I assume "switch" means change the places of the two coins, not flip them in place.)

A global view of this problem starts with the observation that the system has $8$ distinguishable states: the positions of each of the two coins and which side of each is up are independent binary choices.

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  • $\begingroup$ "switch" means change the places of the two coins, correct $\endgroup$
    – nz_
    Commented May 18 at 14:53
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    $\begingroup$ ah... I was assuming coins are indistinguishable from each other $\endgroup$
    – nz_
    Commented May 18 at 14:59
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    $\begingroup$ Use a dime and a quarter! $\endgroup$
    – Lee Mosher
    Commented May 18 at 15:10
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    $\begingroup$ I really think the intention of the question is for the coins to be identical, and the only information needed to describe a state is what is showing (Heads or Tails) in each location. Regardless, this extra information (which coin is which) is not needed to distinguish $M_3$ and $M_7$. For both interpretations of the question, the resulting group is dihedral of order $8$. $\endgroup$
    – tkf
    Commented May 19 at 4:18
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If the coins are distinguishable, then $M_3$ and $M_7$ are different. If they aren't distinguishable, then switching the two coins is the same as switching the two states (e.g. if you have a Heads in Location A and a Tails in Location B, then $M_4$ is the same as making the coin in Location A Tails and making the coin in Location B Heads).

Also, according to the nomenclature I am familiar with, * is not an "operation" but an "operator". I would call the elements of $G$ operations, as they are acting on the set of coins, while * is an operator, as it acts on the elements of $G$.

Even if $M_3$ and $M_7$ were the same operation, this is not a problem for the requirement of * being unambiguous. If $M_3$ and $M_7$ are the same, then the symbols "$M_3$" and the symbols "$M_7$" are different sets of symbols, but they are the same element; they are different representatives of the same thing.

You can think of it like two people performing a calculation to find what the output of a function is, and one gets $\frac 12$ as their answer while the other gets $0.5$. This doesn't mean that one of them got the wrong answer, or that the function isn't well-defined. It's just that they have different representations of the same answer.

If you're going to study algebra, you really need to distinguish between a mathematical object, versus a representation of that object.

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    $\begingroup$ If you have Heads in Location A and a Tails in Location B, then the effects of $M_3$ and $M_7$ are different. Thus they represent different elements of the group. $\endgroup$
    – tkf
    Commented May 19 at 3:56
  • $\begingroup$ that's a great glimpse / point you highlighted regarding the operation if the coins were indistinguishable. i.e. even though $M_3$ and $M_7$ are different symbols, they would be the same element $\endgroup$
    – nz_
    Commented May 19 at 20:26

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