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.

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

{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