How to find the number of elements with a particular order in a group?

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SUMMARY

This discussion focuses on determining the number of elements with a specific order in symmetric groups, particularly Sn. The process involves decomposing elements into disjoint cycles, calculating the number of irreducible cycles of a given length, and summing all possibilities to arrive at the final count. A key point raised is the absence of a subgroup of order 6 in S4, leading to the conclusion that no elements of order 6 exist in S4, which raises questions about the validity of the initial assumptions regarding element orders.

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  • Understanding of symmetric groups, specifically Sn
  • Knowledge of cycle decomposition in group theory
  • Familiarity with the concept of least common multiples in relation to cycle orders
  • Ability to calculate permutations and combinations
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  • Study the properties of symmetric groups, focusing on Sn and its elements
  • Learn about cycle decomposition and its applications in group theory
  • Explore the relationship between subgroup orders and element orders in finite groups
  • Investigate the counting techniques for permutations, particularly in relation to cycle types
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Students preparing for exams in abstract algebra, mathematicians studying group theory, and educators teaching concepts related to symmetric groups and cycle structures.

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I'm preparing for an upcoming exam, and as I see one of the typical questions that is frequently asked in our exams is about finding the number of elements that have a particular order in a group like Sn.

I searched on google and came up with some such problems with solutions. To be honest, there are still points that I don't understand well. I think if such a problem is given, I have to follow a path like this:

Assume that I want to find all elements of order m in Sn.

1. First of all, I have to list the ways that I can decompose an element of order m into disjoint cycles. I know that for two disjoint cycles of order s and t the order of their composition will be the least common multiple of s and t. So, the first natural question that arise is how I should know how many such decompositions are there and how I can list them without possible mistakes.

2. Then I must find the number of ways I can write an irreducible cycle of length t. For example if I want to count the number of ways we can form a 4-cycle as (a b c d) in S6 I should say that we have 6 choices for the first place, 5 choices left for the second place, and so on... and then we have to divide by 4 because we have considered (a b c d), (b c d a), (c d a b) and (d a b c) as different cycles.

3. Now I sum all the possibilities to get the final answer.

But I still don't get it. I mean if I remember it correctly there is no subgroup of order 6 in S4. Doesn't it mean that there can not exist an element of order 6 in S4 too? If yes, doesn't this contradict what I have said so far?
 
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I don't really get it:

math.geek said:
I'm preparing for an upcoming exam, and as I see one of the typical questions that is frequently asked in our exams is about finding the number of elements that have a particular order in a group like Sn.

I searched on google and came up with some such problems with solutions. To be honest, there are still points that I don't understand well. I think if such a problem is given, I have to follow a path like this:

Assume that I want to find all elements of order m in Sn.

1. First of all, I have to list the ways that I can decompose an element of order m into disjoint cycles. I know that for two disjoint cycles of order s and t the order of their composition will be the least common multiple of s and t. So, the first natural question that arise is how I should know how many such decompositions are there and how I can list them without possible mistakes.

2. Then I must find the number of ways I can write an irreducible cycle of length t. For example if I want to count the number of ways we can form a 4-cycle as (a b c d) in S6 I should say that we have 6 choices for the first place, 5 choices left for the second place, and so on... and then we have to divide by 4 because we have considered (a b c d), (b c d a), (c d a b) and (d a b c) as different cycles.

3. Now I sum all the possibilities to get the final answer.

Here you are talking about elements of order 4 in ##S_6##, and

But I still don't get it. I mean if I remember it correctly there is no subgroup of order 6 in S4. Doesn't it mean that there can not exist an element of order 6 in S4 too? If yes, doesn't this contradict what I have said so far?

Here you are talking about elements of order 6 in ##S_4##.
 

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