# Order of this Group

1. Mar 12, 2016

### RJLiberator

1. The problem statement, all variables and given/known data
Let f,h ∈S_4 be described by:
f(1) = 1
f(2) = 4
f(3) = 3
f(4) = 2

h(1) = 4
h(2) = 3
h(3) = 2
h(4) = 1

Express (f°h) in terms of its behavior on {1,2,3,4} and then find the order of (f°h).

2. Relevant equations

3. The attempt at a solution

So, first I express (f°h) in terms of its behavior.
(f°h)(1) = 2
(f°h)(2) = 3
(f°h)(3) = 4
(f°h)(4) = 1

Done!
Now, find the order of (f°h):

The order of (f°h) is min{K:g^k-e} or infinity if no such k exists.

I'm having a bit of difficulty in this. To me, when I search wikipedia: https://en.wikipedia.org/wiki/Order_(group_theory)
It says that the order is the cardinality or the number of elements in the set.

(f°h)'s order thus is 4?
But I have a feeling it might be 3 based on in class examples.
My answer is 4.

How did I do?

2. Mar 12, 2016

### Orodruin

Staff Emeritus
You need to separate the order of a group, i.e., the number of elements it contains, from the order of a group element, i.e., the power you need to raise it to to get the identity element.

I would also suggest using cycle notation for group elements of $S_n$, it reduces what you have to write significantly.

3. Mar 12, 2016

### RJLiberator

So the number of elements it contains is 4.
The order of a group element (the power needed to raise it to get the identity element) is 1?
so 4-1 = 3?

4. Mar 12, 2016

### Orodruin

Staff Emeritus
No, the point is not to mix the concepts. The group is $S_4$ and its order is $4!$, but that was not what the question was about. The question is about the order of the group element fh.

5. Mar 12, 2016

### RJLiberator

Ah. So that is the distinction.
I'm sorry, I made a typo in the question f,h was supposed to be (f°h)

The order of the group element (f°h) is thus 1 as any element in (f°h) has a power of 1.

6. Mar 12, 2016

### Orodruin

Staff Emeritus
No. The order of a group element $f$ is the lowest number $k$ such that $f^k = e$, where $e$ is the identity element (which is the only element of order one!).

7. Mar 12, 2016

### RJLiberator

Hm.
OH. The order is 4? Because (f°h)(4) = 1!

That's the e, the identity.

I think the light went off in my head. Correct?

8. Mar 12, 2016

### Orodruin

Staff Emeritus
The order is 4, but the reason is not the one you stated. Try to figure out the order of f and h to start with. Note that also h(4)=1.
Edit: Are you familiar with cycle notation? It will simplify your life significantly.

9. Mar 12, 2016

### RJLiberator

Damn.

The order of h is 4.
The order of f is 1.

f composed of h is thus 4*1 = 4 ?

10. Mar 12, 2016

### Orodruin

Staff Emeritus
No. What is h^2?

11. Mar 12, 2016

### Orodruin

Staff Emeritus
And again note that the only element with order one is the identity, for no other element $f$ is $f^1=f=e$.

12. Mar 12, 2016

### RJLiberator

So it deals with powers ?

h^2 = I am not even really sure how to express this. Perhaps this is my problem.

h^2 = h(2)^2 = 9 ?

13. Mar 12, 2016

### Orodruin

Staff Emeritus
No, h^2 is the usual notation for h×h, where × is the group operation.

14. Mar 12, 2016

### RJLiberator

Ok.
h^2 then is

(h°h)(1) = 1
(h°h)(2) = 2
(h°h)(3) = 3
(h°h)(4) = 4

(f°f) is similar.

15. Mar 12, 2016

### Orodruin

Staff Emeritus
Right, so what is the order of f and h? If you apply the same reasoning to f×h, what is its order?

16. Mar 12, 2016

### RJLiberator

17. Mar 12, 2016

### Orodruin

Staff Emeritus
No, taking a power of the group operation has nothing to do with taking a power of the elements you rearrange. Still, I suggest you first figure out the orders of f and h.

18. Mar 12, 2016

### RJLiberator

Oh, it deals with permutations (from reading the link?)

{1, 2, 3, 4} = the permutations are thus for the power of say, 2 we get {(1,3)(2,4)}
And so the permutations of 4 is the identity back.

19. Mar 12, 2016

### Orodruin

Staff Emeritus
So what are the orders of the elements f, h, and f×h?

20. Mar 12, 2016

### RJLiberator

f = {1, 4, 3, 2}
so the order is 3
due to permutations

h = {4, 3, 2, 1}
so the order is 4

fxh = { 2, 3, 4, 1}
so the order is 4

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