Proving the Order of the Symmetric Group S_N: Where to Begin?

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SUMMARY

The symmetric group S_N, representing all permutations of N elements, has an order of N!, which can be proven through a constructive approach. For any set of N elements, the first element has N choices for mapping, the second has N-1 choices, and this pattern continues until the last element, leading to the conclusion that the total number of permutations is the product N × (N-1) × ... × 1, which equals N!. This foundational concept is crucial in group theory and combinatorics.

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  • Understanding of group theory concepts, particularly symmetric groups.
  • Familiarity with permutations and bijections.
  • Basic knowledge of factorial notation and its properties.
  • Experience with mathematical proofs and logical reasoning.
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  • Study the properties of symmetric groups and their applications in combinatorics.
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  • Investigate related topics such as the alternating group A_N and its relationship to S_N.
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Homework Statement



Show that the symmetric group (permutation group) S_N is of order N!

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The Attempt at a Solution



I can't get started on how to prove this. I understand that if

n=2 S_2= {E, (12)}
n=3 S_3={E, (12), (13), (23), (123), (132)}

and so on. This makes this seem kind of intuitive. However I can't even get started on proving it for N. Could somebody please help me get started?
 
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Given {1, 2, ..., n-1, n}, you want to construct a bijection onto itself. So take the first element, 1. There are n choices of elements it can map to. Then consider 2. There are n-1 elements it can map to (since the function is 1-1 and one of the n elements is already 'taken' by 1)
 
Got it. Thanks.
 

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