How many elements of order 2 are contained in S_4?

  • Thread starter Thread starter Samuelb88
  • Start date Start date
  • Tags Tags
    Elements
Click For Summary

Homework Help Overview

The discussion revolves around determining the number of elements of order 2 in the symmetric group S_4, which consists of all permutations of four elements.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the nature of elements in S_4, particularly focusing on transpositions and their orders. Questions arise regarding the completeness of counting elements of order 2, including whether other forms of permutations contribute to this count.

Discussion Status

There is an ongoing exploration of the types of permutations that qualify as elements of order 2. Some participants have identified transpositions, while others have pointed out the existence of additional permutations, such as products of disjoint transpositions. The discussion reflects a mix of agreement and uncertainty about the total count.

Contextual Notes

Participants reference the total number of elements in S_4 and engage in clarifying the definitions and properties of permutations, particularly in relation to their orders. There is also mention of a useful external resource for further exploration.

Samuelb88
Messages
160
Reaction score
0

Homework Statement


How many elements of order 2 does the symmetric group S_4 contain?


Homework Equations





The Attempt at a Solution


I know that transpositions have order two. Moreover, any k-cycle has order k. Thus there are six elements with order two contained in S_4?

Is this all? Seems too simple.
 
Physics news on Phys.org
How many elements are there in S4 and what are they?

ehild
 
ehild,

There are 4! = 24 elements in S_4. The elements are all such permutations of {1,2,3,4}.

Here is how I see the problem:

Let \sigma \in S_4. Then either:
1. \sigma = (a_1 a_2) and (a_1 a_2)^2 = e \Rightarrow order 2.
2. \sigma = (a_1 a_2 a_3 and (a_1 a_2 a_3)^3 =e \Rightarrow order 3.
3. \sigma = (a_1 a_2 a_3 a_4) and (a_1 a_2 a_3 a_4)^4 = e \Rightarrow order 4.

Thus there are 6 permutations of order 2 in S_4. Is this not correct?
 
Last edited:
Oh, you mean the permutation group? I thought it was the point group called S4 - sorry.

ehild
 
Samuelb88 said:
Thus there are 6 permutations of order 2 in S_4. Is this not correct?

No, this is not correct.
Did you find how many of order 3 and 4 there are?
Do they add up to 24?
 
well, I got an idea but am not sure if I'm right because I haven't had abstract algebra yet. well, what you're claiming is that in a cyclic group G an element in G is a transposition if and only if It is of order 2. well, It's obvious that any transposition element in G is of order 2 but can we say that any element of order 2 is a transposition?
if yes, then your question would become that in how many ways we can permute two letters from n letters keeping the others the same position they are. That would be an easy problem in combinatorics and discrete math.
 
Hi Samuelb88! :smile:

You are certainly correct that there are 6 transpositions in S4, i.e. there are 6 elements of the form (a b). These are

(1~2),(1~3),(1~4),(2~3),(2~4),(3~4)

However, these are not the only elements of order 2! For example

(1~2)(3~4)

is also of order 2, so you got to count this one too!
 
Oh right! So that means there are six transpositions and three disjoint transpositions in S_4[\itex].
 
  • #10
Yes, that sounds right!
 
  • #11
Thanks guys!
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
  • · Replies 29 ·
Replies
29
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K