Permutations of a single number in the symmetric group

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In the symmetric group S_5, the permutations of the set {2,5} include the identity e and the transposition (25). For the set {3}, the permutations are also e and the 1-cycle (3), which is considered the same as e in terms of identity. It's clarified that 1-cycles do not need to be explicitly written in permutations, as they do not change the overall permutation. The total number of permutations for n elements is n!, confirming that for 2 elements there are two permutations and for 1 element, there is one.
Ted123
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Say we have the symmetric group S_5.

The permutations of \{2,5\} are the identity e and the transposition (25).

But what are all the permutations of \{3\}? Is it e and the 1-cycle (3)?
 
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Ted123 said:
Say we have the symmetric group S_5.

The permutations of \{2,5\} are the identity e and the transposition (25).

But what are all the permutations of \{3\}? Is it e and the 1-cycle (3)?

What do you think would be the difference between e and (3)? Aren't they both the identity permutation?
 
Dick said:
What do you think would be the difference between e and (3)? Aren't they both the identity permutation?

Oh yeah of course. You don't have to write 1-cycles in a permutation so e=(1)(2)(3)(4)(5)=(1)=(2)=(3)=(4)=(5)=(1)(2)=(1)(3) etc.
 
Ted123 said:
Oh yeah of course. You don't have to write 1-cycles in a permutation so e=(1)(2)(3)(4)(5)=(1)=(2)=(3)=(4)=(5)=(1)(2)=(1)(3) etc.

Right. The number of permutations on n elements is n!. So for 2 elements you have two permutations, for 1 element you have one.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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