How many elements of order 5 are in S7?

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SUMMARY

The number of elements of order 5 in the symmetric group S7 can be determined by understanding the structure of permutations. In S7, a cycle with 5 elements has an order of 5, and the product of two disjoint cycles of lengths p and q has an order equal to the least common multiple (lcm) of p and q. Since S7 consists of all permutations of 7 objects, the relevant cycles can be expressed as products of disjoint cycles, such as (12345)(67) for elements of order 5. This analysis confirms that permutations in S7 can be represented as cycles, but not all permutations form cyclic groups.

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  • Understanding of symmetric groups, specifically S7
  • Knowledge of cycle notation in permutations
  • Familiarity with the concept of order in group theory
  • Basic understanding of least common multiples (lcm)
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  • Study the structure of symmetric groups, focusing on S_n
  • Learn about cycle decomposition of permutations in group theory
  • Explore the properties of cyclic groups and their generators
  • Investigate the application of lcm in determining orders of permutations
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Mathematicians, students of abstract algebra, and anyone studying group theory, particularly those interested in permutations and symmetric groups.

williamaholm
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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:
 
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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.
 
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:
 

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