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## Main Question or Discussion Point

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 S

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 S

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 S

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 S

_{n}.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 S

_{n}.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 S

_{6}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 S

_{4}. Doesn't it mean that there can not exist an element of order 6 in S_{4}too? If yes, doesn't this contradict what I have said so far?