# Permutation describing action on G-set

1. Apr 23, 2012

### Jesssa

hey,

i have a quick question about G-sets and how the action of an element of the group on the set can be represented as a permutation of the elements of the set.

If you had a set S = {1,2,..,50} and a permutation σ: S→S where |σ|=30,

Does this mean that the minimum number of fixed points σ has is 50 - 30 = 20?

Since only 30 of the elements in the set are effected by the permutation?

2. Apr 23, 2012

### I like Serena

Hi Jesssa!

|σ|=30 means that the order of σ is 30.
This is already the case when for instance σ=(12)(345)(6789 10).
But dividing up all elements in such cycles (with lengths that divide 30) will also yield a permutation of order 30.

3. Apr 23, 2012

### Jesssa

Hey I like Serena,

But each rearrangement dealt by the permutation of order 30 acting on the G-set S, (which has order 50) only rearranges 30 of the 50 elements of S.

So under the action only 30 out of 50 of the elements are effected and hence rearranged by the permutation leaving the remaining 20 elements fixed right?

Fixed as in, under the action they remain themselves.

If you were to have a permutation of order 30 σ=(12)(345)(6789 10), that wouldn't provide the minimum number of fixed points right?

That's saying that 10 of the elements are rearranged and the rest remain fixed
σ=(12)(345)(6789 10)(11)(12)(13)...(50)

So is it true that the minimum number of fixed points is 20? because the maximum length of σ is 30, which rearranges 30 elements leaving 20 fixed?

If it were made up of cycles which divide 30, it would describe a action which effects less and less of the elements in the set, increasing the number of fixed elements?

4. Apr 23, 2012

### DonAntonio

Once again, no. The permutation $(1\,2\,3 .... 29\, 30)(31 \,32)(33 \,34)....(49\, 50)$ also has order 30 and it has zero (0) fixed points...

DonAntonio

5. Apr 23, 2012

### I like Serena

The order of an element is unrelated (more or less) to the order of the set S.
The order of a set S is the number of elements.
The order of an element is the minimum number of times you need to apply that element to itself before you get the identity element.
Btw, it is true that the order of an element always divides the order of the set it works on.

In the example I gave, σ=(12)(345)(6789 10),
which is the same as σ=(12)(345)(6789 10)(11)(12)(13)...(50),
10 elements are shuffled and 40 remain fixed.
But it takes 30 repeated applications of σ before you get identity.

So the minimum number fixed is zero (as DonAntonio's example does).
And the maximum number fixed is 40 (as my example does).

6. Apr 23, 2012

### Jesssa

Thanks guys

I just made up another one to see if I get it properly,

if X = {1,2...,10001} and the order of the permutation σ is 49,

then you can do the same thing as DonAntonio did,

(1 2 3 .... 49)(50 ... 57)...in groups of 7, up to 9947

this keeps the order of the permutation 49

Since 10001-49 = 9952 and 9952/7 is 1412 with remainder 5,

so the minimum number of fixed points has to be 5?

Last edited: Apr 23, 2012
7. Apr 23, 2012

### DonAntonio

In fact it is up to 9,948, as any 7-cycle in your example begins and ends with an

integer = 1 (mod 7). All the rest looks fine to me.

DonAntonio

8. Apr 23, 2012

### Jesssa

Cool!

Thanks a lot!

9. Apr 29, 2012

### Jesssa

Hey DonAntonio,

if X = {1,2...,10001} and the order of the permutation σ is 49,

the way I wrote it before was

(1 2 3 .... 49)(50 ... 57)...(9941...9948)

but looking over it again makes it look like the elements in positions 9949 to 9996 are also fixed because this permutation does not permute them, also (50 ... 57)...(9941...9948) are chains of length 8 not 7, eg (50 51 52 53 54 55 56 57)

is that correct?

I think this is incorrect because from what I remember I thought 9948 was the correct ending number because 9952/7 = 1421 remainder 5, and 7x1421 = 9947 and as you said it is actually up to 9948.

But the reason I actually did 10001 - 49 = 9952 then 9952/7 was to determine the number of 7 length cycles would be after the 49 length cycle

Does this mean the correct way to write it is
σ= (1 2 3 .... 49)(50...56)(57...63).....(9990....9996)

or another way using cycles of length 49
=(1 2 3 .... 49)(50....98)....(9948...9996)

and both of these permutations have only 5 fixed elements,

Last edited: Apr 29, 2012
10. Apr 29, 2012

### DonAntonio

Yeah, well: the one important thing, imo, is to keep in mind that one can construct ANY 49-cycle and then multiply by other 7-cycles

or 49-cycles and the outcome is a permutation of order 49. How to do it is pretty ugly and cumbersome and I leave that to students (and to you).

DonAntonio

Pd. BTW, next time I think it'd be better to start a new thread.