How many elements of order 5 are in S7?

  • Thread starter williamaholm
  • Start date
  • Tags
    Elements
In summary, the conversation discusses the set S_7, which is the set of all permutations of 7 objects. The main question is how many elements of order 5 are in S_7 and if these elements are also permutations. It is clarified that S_7 is a permutation of order 7 and not 5. The concept of cyclic groups is also brought up, with the explanation that permutations are not cyclic groups but can generate cyclic subgroups. The conversation also mentions that permutations can be written as cycles and that a cycle with n elements has an order of n. The product of two disjoint cycles is also discussed and it is mentioned that this information can help find all the elements of S_7 with order 5. One
  • #1
williamaholm
3
0
Greetings, I just stumbled upon this site and thought someone might be able to help me. How many elements of order 5 are in S7? Are the elements permutations? I thought that S7 was a permutation of order 7. Also, are permutations cyclic groups? I know that permutations are a group and can be written as cycles ,but is this "cycle" the same as a "cyclic" group? Do these questions make any sense? :confused:
 
Physics news on Phys.org
  • #2
S_7 is the set of all permutations of 7 objects. I've never seen S7 written to be honest, they tend to use subscripts.

'are permutations cyclic groups' is no, since one is an element of a group, and one is a group.

A cyclic group is one generated by a single element.

Given a cycle in S_n it generates a cyclic subgroup, which is a proper subgroup unless n=2.

In usual notation, a cycle is something of the form (abc...d), and any permutation is writable as a product of disjoint cycles, that is cycles that have no common elements. eg (123)(45) is a product of disjoint cycles, but (12)(23) isn't. It is equal to (123).

A cycle with n elements in it has order n.

The product of two disjoint cycles of lengths p and q has order lcm(p,q)

this is enough to find all the elements of S_7 that have order 5.
 
  • #3
Thank you, matt! your answer is really helpful for my present study! I hope people would feel the same..:niceteeth:
Oh, sorry so unbearable that I have to :biggrin:
 

1. How many elements of order 5 are in S7?

There are 120 elements of order 5 in S7. This can be calculated by using the formula n!/r, where n is the total number of elements (7) and r is the order of the element (5).

2. How do you determine the order of an element in S7?

The order of an element in S7 can be determined by finding the smallest exponent of the element that results in the identity element. For example, if an element raised to the power of 5 results in the identity element, then the order of that element is 5.

3. What is the identity element in S7?

The identity element in S7 is the permutation that leaves all elements unchanged. In other words, it is the element that when composed with any other element, results in the same element.

4. Is S7 a finite or infinite group?

S7 is a finite group, as it contains a finite number of elements (5040 to be exact).

5. How many elements of order 1 are in S7?

There is only one element of order 1 in S7, which is the identity element. This is because any element raised to the power of 1 results in the same element, so all other elements in S7 have a higher order.

Similar threads

  • Linear and Abstract Algebra
Replies
7
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
2K
  • General Math
Replies
1
Views
732
Replies
2
Views
973
  • Linear and Abstract Algebra
Replies
3
Views
970
  • Linear and Abstract Algebra
2
Replies
44
Views
5K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Linear and Abstract Algebra
Replies
2
Views
993
Back
Top