# Permutation Loops- successor operation

1. Jul 1, 2005

### MathematicalPhysicist

in this page ive encountered this topic (it's a topic from combinatorics, so it's relevant to discrete maths with sets and so on [that's my justification to post it here ), anyway from my point of view the page describes poorly the successor operation:
"The resulting
sequence of elements constitutes another loop, which we will call
the "successor" of the original loop."

here's an example of such successor from the webpage:
I = 123 C = 231
A = 132 D = 312
B = 213 E = 321
IDEACB = A
DEACBI = A
EACBID = B
ACBIDE = A
CBIDEA = A
BIDEAC = E
i dont understand why for example the first sequence of elements equals A (or is identical to it), perhaps i missed something from the explanation, anyone care to explain.

btw, here's the page:
http://www.mathpages.com/home/kmath031.htm

2. Jul 1, 2005

### AKG

There is a better notation than this. When they say I = 123, they mean that I sends 1 to 1, 2 to 2, and 3 to 3. If you have X = xyz, then X sends 1 to x, 2 to y, and 3 to z. A more standard notation would be to write it like this:

Code (Text):
X = [1 2 3]
[x y z]
So X sends whatever is in the top row to the thing directly above it. So in particular, we have:

Code (Text):
A = [1 2 3]
[1 3 2]
A basically just swaps 2 and 3. This is commonly denoted as (23). We will use this notation from now on. X = (xy) means that it sends x to y, y to x, and leaves z fixed. X' = (x'y'z') means that it sends x' to y', y' to z', and z' to y'. We can denote I as (), since it does nothing.

Now, if you have something like:

(123)(23)(23)(132)(231)(12)(321)

then you read it from right to left. The right-most one sends 1 to 3. The next one leaves 3 fixed. The next one sends 3 to 1, and then 1 to 3, 3 to 2, 2 to 3, and finally 3 to 1, so this whole thing sends 1 to 1. Next, look at 2. The right most one sends 2 to 1, then the next one sends 1 to 2, the next one sends 2 to 3, then 3 to 2, then 2 to 3, then 3 to 2, then 2 to 3, so this whole thing sends 2 to 3. Obviously, it sends 3 to 2, so we get:

(123)(23)(23)(132)(231)(12)(321) = (23) = (32)

Note that (23) and (32) are the same. However, given your original notation, 23 and 32 are not even defined. You should also notice that (123) and (312) are the same, but given your original notation, 123 and 312 are defined but are not equal, since 123 is:

[1 2 3]
[1 2 3]

which is just (), and 312 is:

[1 2 3]
[3 1 2]

which is (132).

IDEACB
= ()(132)(13)(23)(123)(12)
= (23)
= A

as required. See if you can figure out the rest for yourself.

3. Jul 1, 2005

### AKG

By the way, I don't think the site did a bad job of explaining things, it seems that you're just unfamiliar with the permuation groups. Do you know what a group is? As far as I can tell, this has much more to do with group theory (part of abstract algebra) and much less to do with combinatorics. There is some relation to the idea of permutations as a group, and permutations where you calculate things like "n arrange k," but from the part of that link that I read, this has much less to do with calculating things like "n arrange k" and more to do with group theory. At very least, it seems to require some knowledge of group theory but no particular knowledge of combinatorics. Maybe further down the page on that link, things are different...

4. Jul 2, 2005

### MathematicalPhysicist

ok, i understand it, a bit tricky i think but nice. anyway my knowledge on group theory mainly stands on knowledge of the axioms (and when the author did specify accosiativity and identity element it felt as if it's connected to groups), and the simple theorems with the identity element of a group, not a lot i know

btw, thanks for the help.