What is the explanation for the permutation table in my book?

[STEMucator]

Homework Statement

This is not really a question. More of an understanding issue. They have created a table for the alternating group of degree 4 in my book and I'm having trouble understanding it.

http://gyazo.com/87980bef3669619796dda2f55a9d04d1

Homework Equations

Cycle notation.

The Attempt at a Solution

So I can understand why (1)(1) = (1), but for something like (12)(34)(12)(34) = (1) I am unsure of how they've done this and then later on they somehow get (123)(123) = (13)(12)(13)(12) = (132).

The table in general is somewhat confusing to me. Any help with this one would be appreciated.

 

[jbunniii]

but for something like (12)(34)(12)(34) = (1) I am unsure of how they've done this

Just consider what the map does to each of 1, 2, 3, and 4. Start with 1. I assume you're composing from right to left, so (3 4) is the first cycle to be applied. This does not change 1. Thus (3 4) maps 1 to 1. Next, (1 2) maps 1 to 2. Then (3 4) leaves alone. Finally (1 2) moves 2 back to 1. Therefore the composition of the 4 maps has left 1 unchanged. You can verify for yourself that the same is true for 2, 3, and 4.

and then later on they somehow get (123)(123) = (13)(12)(13)(12) = (132).

Once again, to verify equality, just check what each of the three maps does to 1, 2, and 3.

The first map is (1 2 3)(1 2 3). The first (right) cycle maps 1 to 2, and then the second (left) cycle maps 2 to 3. So the composition maps 1 to 3. Similarly, the composition maps 2 to 1, and 3 to 2. Therefore it is equal to the cycle (1 3 2). You can confirm that the middle map is equal to this same cycle.

 

[STEMucator]

Just consider what the map does to each of 1, 2, 3, and 4. Start with 1. I assume you're composing from right to left, so (3 4) is the first cycle to be applied. This does not change 1. Thus (3 4) maps 1 to 1. Next, (1 2) maps 1 to 2. Then (3 4) leaves alone. Finally (1 2) moves 2 back to 1. Therefore the composition of the 4 maps has left 1 unchanged. You can verify for yourself that the same is true for 2, 3, and 4.



Once again, to verify equality, just check what each of the three maps does to 1, 2, and 3.

The first map is (1 2 3)(1 2 3). The first (right) cycle maps 1 to 2, and then the second (left) cycle maps 2 to 3. So the composition maps 1 to 3. Similarly, the composition maps 2 to 1, and 3 to 2. Therefore it is equal to the cycle (1 3 2). You can confirm that the middle map is equal to this same cycle.

Ahhhh I understand what's happening now.

So for example, let's take a harder one.. say (14)(23)(234) = (143).

This happens because 1 gets mapped to 4, 2 gets sent to itself, 3 gets sent to 4 which gets sent back to 1 and finally 4 gets sent to 2 which gets sent to 3.

Thus we get the cycle (143)(2) = (143).

Thank you for clearing this up for me, I was having trouble reading from right to left for some reason.

 

[jbunniii]

Ahhhh I understand what's happening now.

So for example, let's take a harder one.. say (14)(23)(234) = (143).

This happens because 1 gets mapped to 4, 2 gets sent to itself, 3 gets sent to 4 which gets sent back to 1 and finally 4 gets sent to 2 which gets sent to 3.

Thus we get the cycle (143)(2) = (143).

Yes, that's correct. Reading right-to-left can be a bit confusing at first, but it makes sense because it works the same way for functions: f \circ g \circ h means apply h, then g, then f.

Don't worry, you'll get used to it and it will become second nature, and then suddenly you will take some higher level group theory course from an old-fashioned instructor who insists on writing everything left-to-right and accordingly writes mappings on the right, i.e. what most people would call f(g(h(x))) suddenly becomes (((x)h)g)f. I wish I was joking about this, but I've had such instructors (and textbooks).