1. ## Coin Game

Hi.

Imagine 2 coins on a table, at positions A and B. In the game there are 8 possible moves:

$M_1:$ flip the coin at A.
$M_2:$ flip the coin at B.
$M_3:$ flip 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, then switch.
$I:$ do nothing.

We may consider an operation on the set $\{I,M_1,...,M_7\}$, which consists of performing any 2 moves in succession. EG: If we switch coins, then flip at A, this is the same as first flipping over the coin at B then switching.

Write the table for $(G,*)$.

Ok, so I understand the notion of group, I understand how to create operation tables, but I'm having trouble visualizing the equivalences between a pair of elements and the associated element which is the pair's product. For example,

$M_1*M_2=?$.

Can anyone help me see this?

*edit>
OK. So, what about $M_5*M_2$ ? The way that I think about this is like, if we let the coins have sides 1 and 2 at points A and B, respectively, then if side 1 is showing at both A and B, and $M_5$ is made, then 1 becomes 2 at A and then moves to B. So, we have 1 at A and 2 at B. Now, we do $M_2$ and 1 now shows at B. This is where we started. So, is $M_5*M_2=I$ or $M_5*M_2=M_4$. To me it seems as if $*$ is not uniquely defined.

2. ## Re: Coin Game

Originally Posted by VonNemo19
Hi.

Imagine 2 coins on a table, at positions A and B. In the game there are 8 possible moves:

$M_1:$ flip the coin at A.
$M_2:$ flip the coin at B.
$M_3:$ flip 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, then switch.
$I:$ do nothing.

We may consider an operation on the set $\{I,M_1,...,M_7\}$, which consists of performing any 2 moves in succession. EG: If we switch coins, then flip at A, this is the same as first flipping over the coin at B then switching.

Write the table for $(G,*)$.

Ok, so I understand the notion of group, I understand how to create operation tables, but I'm having trouble visualizing the equivalences between a pair of elements and the associated element which is the pair's product. For example,

$M_1*M_2=?$.

Can anyone help me see this?

*edit>
OK. So, what about $M_5*M_2$ ? The way that I think about this is like, if we let the coins have sides 1 and 2 at points A and B, respectively, then if side 1 is showing at both A and B, and $M_5$ is made, then 1 becomes 2 at A and then moves to B. So, we have 1 at A and 2 at B. Now, we do $M_2$ and 1 now shows at B. This is where we started. So, is $M_5*M_2=I$ or $M_5*M_2=M_4$. To me it seems as if $*$ is not uniquely defined.
The reason you think that * is not uniquely defined is that you are treating the coins as indistinguishable, which is not the case. (If it were, then $M_4$ would be the same as $I.$)

To keep track of the situation, you need to keep track of which coin is in which position, as well as which side of each coin is showing. Suppose you call the coins x and y. Use the notation $x_1$ to denote that coin x is showing side 1, and similarly $x_2,$ $y_1$ and $y_2.$

Suppose that the system starts in the state $(x_1,y_1),$ which means that coin x is at A (showing side 1) and coin y is at B (also showing side 1). After the move $M_5$ is made, the system will be in state $(y_1,x_2).$ If you then make the move $M_2,$ the state will become $(y_1,x_1).$ So both coins show side 1, but their positions have switched. Thus $M_5*M_2 = M_4,$ and the ambiguity that troubled you has been resolved.

3. ## Re: Coin Game

Opalg. I'm honored that I've finally reached the point where MHF's best mathematician has found my question interesting enough to respond to. Thank you.

You've helped a lot. I've decided to continue through this elementary treatment of algebra at a snail's pace so that I don't miss a thing. Therefore, I will have many more questions on the topic.

C-Ya