# A coin game - is it a group?

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)?

• $M_1 * M_2 \neq M_7$: $(M_1 * M_2)(H, T) = M_1(H, H) = (T, H) \neq (H, T) = M_7(H, T)$ (writing $H$ for heads and $T$ for tails). Commented May 18 at 14:51
• 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 $I$. Commented May 19 at 14:55
• @RossMillikan A state of the game does not need to include which coin is where - only what the coins at each location are showing.
– tkf
Commented May 19 at 17:29

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:

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.

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 $$8$$.

• 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).
– nz_
Commented May 19 at 20:57
• 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.
– nz_
Commented May 19 at 20:58
• 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
– nz_
Commented May 19 at 20:59
• 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.
– tkf
Commented May 20 at 0:38

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.

• "switch" means change the places of the two coins, correct
– nz_
Commented May 18 at 14:53
• ah... I was assuming coins are indistinguishable from each other
– nz_
Commented May 18 at 14:59
• Use a dime and a quarter! Commented May 18 at 15:10
• 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 $8$.
– tkf
Commented May 19 at 4:18

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.

• 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.
– tkf
Commented May 19 at 3:56
• 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
– nz_
Commented May 19 at 20:26