# Permutations physics problem

1. Feb 19, 2009

### duki

1. The problem statement, all variables and given/known data

a = (162)(45)
b = (123)(46)
c = (1362)

Find: ab

2. Relevant equations

3. The attempt at a solution

k, so for a i have this:

| 1 2 4 5 6 |
| 6 1 5 4 2 |

and for b i have this:

| 1 2 3 4 6 |
| 2 3 1 6 4 |

so when I started doing the multiplication I got to '4 -> 5, 5 -> ??'
What am I doing wrong?

here's what I have so far:

| 1 2 4 5 6 |
| 4 2 |

2. Feb 19, 2009

### mXSCNT

Re: Permutations

Work from right to left. First 4 goes to 6 by b, then 6 goes to 2 by a.

3. Feb 19, 2009

### MathematicalPhysicist

Re: Permutations

You need to follow the chart and think of permutation as replacing one term with another.
so you have:
b=| 1 2 3 4 6 |
| 2 3 1 6 4 |
a=| 1 2 4 5 6 |
| 6 1 5 4 2 |

If you need a way to sort this, then write down as ab(5)=a(b(5))=a(5)=4.

4. Feb 20, 2009

### duki

Re: Permutations

I'm not sure I understand... by going from b -> a I would be doing ba not ab, which aren't equivalent?

am i missing something?

5. Feb 20, 2009

### lanedance

Re: Permutations

doesn't it cycle through the numbers...
ie (162) means (1) goes to position (2), then position (6) goes to (1), and (2) to (6)

So with all six number in the chart
and for a = a_II.a_I = (1,6,2)(4,5) apply right hand permutation first gives

|1 2 3 4 5 6|
apply a_I = (4,5)
|1 2 3 5 4 6|
apply a_II = (1,6,2)
|6 1 3 5 4 2|

I don't think permutations necessarily commute, think of (1,2) and (2,3), but try it and see

a = a_II.a_I is actually a set of 2 permutations which do commute, so it doens't matter which is done first

for a.b i would apply b first then a
so b.a = (162)(45)(123)(46)

To write out the result simply, start with where (1) has moves & track each number as it moves through the set from, when you move a number back to (1) that ends that permutation chain & look at the remaining numbers.

Last edited: Feb 20, 2009
6. Feb 20, 2009

### Tobias Funke

Re: Permutations

Are you working in S_6={1,2,3,4,5,6}? If you know the set being permuted, then 1-cycles aren't included in cycle notation. a takes 3 to 3, which is understood since 3 isn't included in a cycle. You would write a as
| 1 2 3 4 5 6 |
| 6 1 3 5 4 2 |

Similarly, b takes 5 to 5. It seems like not including these is giving you problems. Also, check your book to see if (ab)(x) is defined as a(b(x)) or b(a(x)). I've only seen it defined as a(b(x)), which is why everyone is telling you to work right to left and do b first, but I think some people define it as b of a.