Proving the Number of Subgroups of Order 6 in S4

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In summary: Sure, you don't have to do ALL of them. Your order 3 subgroups are all conjugate. You really just have to show that one of them is contained in a unique order 6 subgroup, say the one generated by (123). The only 'interesting' 2 cycle to multiply by is (14) (the others will just generate an S3). (123)(14)=(1423). That's good. (123)(1423)=(14246). That's a subgroup of order 6.
  • #1
3029298
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My question:
I was asked to give a subgroup of order 6 of the permutation group S4. That part is not so difficult, for example S3 has order 6 and is a subgroup of S4. But now I have to show how many subgroups of order 6 are in S4. Intuitively thinking, there are four of them, each of them leaving 1, 2, 3 or 4 fixed. But how can you prove this?
 
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  • #2
A subgroup of order 6 must contain an element of order 3 and an element of order 2. That means it contains a 3 cycle and a 2 cycle or a product of two disjoint 2 cycles. Can you show it's a 2 cycle? Let's pick the 3 cycle to be (1,2,3) (how many other choices are there that lead to different order 3 subgroups?). Now you need to add a 2 cycle. If you add a 2 cycle chosen from the elements 1, 2 and 3, like (1,2), you generate S3. If you add an element like (1,4) can you show you get all of S4? Do you see how this proves your conclusion about the 4 different S3's?
 
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  • #3
Hmm, why does the subgroup need to have an element of order 2 and an element of order 3? Why not only elements of order 2 or 3?
 
  • #4
3029298 said:
Hmm, why does the subgroup need to have an element of order 2 and an element of order 3? Why not only elements of order 2 or 3?

Cauchy's theorem says so. 2 and 3 are prime divisors of 6.
 
  • #5
Ah, I see! It cannot be a product of two disjoint 2-cycles because then you again generate S4. And there are four choices of 3-cycles, which gives you 4 subgroups. Is this correct? Thanks a lot :)
 
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  • #6
3029298 said:
Ah, I see! It cannot be a product of two disjoint 2-cycles because then you again generate S4. And there are four choices of 3-cycles, which gives you 4 subgroups. Is this correct? Thanks a lot :)

Right. There are 4 subgroups of order 3. You also have to show there is only one group of order 6 that contains each one.
 
  • #7
Is there an easier way to show that there are no products of disjoint 2-cycles in the subgroup, other than writing all te products with all 3-cycles? (and thus showing that you generate more than 6 elements)
 
  • #8
Sure, you don't have to do ALL of them. Your order 3 subgroups are all conjugate. You really just have to show that one of them is contained in a unique order 6 subgroup, say the one generated by (123). The only 'interesting' 2 cycle to multiply by is (14) (the others will just generate an S3). (123)(14)=(1423). That's not good. It has order 4. No subgroup of order 6 can contain a subgroup of order 4. Stop there. Stuff like that.
 

Related to Proving the Number of Subgroups of Order 6 in S4

1. What is S4 in terms of group theory?

S4 is the symmetric group of order 4, also known as the group of permutations on 4 elements. It consists of all possible ways to arrange the elements 1, 2, 3, and 4 in a row or column.

2. How many subgroups of order 6 can be found in S4?

There are a total of 5 subgroups of order 6 in S4. These include the cyclic subgroups generated by (12)(34), (13)(24), and (14)(23), as well as the two non-cyclic subgroups generated by (123) and (132).

3. How can the number of subgroups of order 6 in S4 be proven?

The number of subgroups of order 6 in S4 can be proven using Lagrange's theorem, which states that the order of a subgroup must divide the order of the group. In this case, the order of S4 is 24, and the only divisors of 24 that are less than 24 are 1, 2, 3, 4, 6, and 8. Therefore, the number of subgroups of order 6 must be one of these numbers. By finding and demonstrating the existence of 5 subgroups of order 6, we can prove that there are no more and that the number is indeed 5.

4. What is the significance of subgroups of order 6 in S4?

In group theory, subgroups of order 6 are important because they are the only subgroups of S4 that are not normal. This means that they are not closed under conjugation, which has important implications for the structure and properties of these subgroups.

5. Are there any other methods to prove the number of subgroups of order 6 in S4?

Yes, there are other methods to prove the number of subgroups of order 6 in S4, such as using the Sylow theorems and group actions. However, using Lagrange's theorem is the most straightforward and commonly used method for this particular problem.

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