If d ≤ n , the symmetric group S n contains the cycle ( 1 , 2 , … , d ) of order d . Additionally, the cycle ( 1 , 2 , … , n ) is an example of an element of order n in S n . Therefore, both conditions are satisfied.
;
S n is the symmetric group of degree n .
If d ≤ n , the cycle ( 1 , 2 , ... , d ) is an element of order d in S n .
The cycle ( 1 , 2 , ... , n ) is an element of order n in S n .
Therefore, S n contains elements of order d if d ≤ n , and it contains an element of order n .
Explanation
Understanding the Problem We want to prove that if d ≤ n , then S n contains elements of order d . Also, we want to find an element of order n in S n for every positive integer n .
Background Knowledge Let's consider the symmetric group S n , which consists of all permutations of n elements. The order of a permutation is the least common multiple (LCM) of the lengths of the cycles in its cycle decomposition.
Proof for the First Part If d ≤ n , we can construct a cycle of length d within S n . Specifically, consider the cycle ( 1 , 2 , ... , d ) . This cycle has length d , and therefore its order is d . Since d ≤ n , this cycle is a valid element of S n .
Finding an Element of Order n Now, let's find an element of order n in S n . We can consider the cycle ( 1 , 2 , ... , n ) . This cycle has length n , so its order is n . This cycle is an element of S n .
Conclusion Therefore, the permutation ( 1 , 2 , ... , n ) is an element of order n in S n .
Examples
Consider arranging n books on a shelf. The symmetric group S n describes all possible arrangements. If you want to arrange the books in a cycle of length d (where d ≤ n ), you can pick d books and arrange them in a cycle. For example, if you have 5 books and want a cycle of length 3, you arrange 3 of them in a cycle, leaving the other 2 fixed. Similarly, to create a full cycle of length n , you arrange all n books in a cyclic order. This concept is used in cryptography, coding theory, and physics to analyze symmetries and patterns.