Show the Symmetric group is generated by the set of transpositions (12) (n-1 n).

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The discussion focuses on proving that the symmetric group S_n can be generated by the set of transpositions {(1 2), (3 4), ..., (n-1 n)}. It begins by establishing that any element in S_n can be expressed as a product of disjoint n-cycles, which can further be represented as products of 2-cycles. The conversation highlights a correction in the initial set notation and emphasizes the need to show that any 2-cycle can be generated from the corrected set. An inductive approach is proposed to demonstrate that any 2-cycle (i j) can be expressed as a product of consecutive transpositions. The discussion concludes with a note on the need for clarity and refinement in the explanation of the proof.
Edellaine
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Homework Statement


5.1: Prove that S_n is generated by the set {(1 2), (3 4),...,(n-1 n)}

Homework Equations


None that I know of

The Attempt at a Solution


Any element in S_n can be written as a product of disjoint n-cycles. So now I need to show any n-cycle can be written as a product of 2-cycles. So if my cycle is (a1 ... ak) = (a1 a2)(a1 a3)...(a1 ak).

So now I've shown S_n can be generated by 2-cycles in general. I'm not sure how to extend this to say that S_n is generated by the set above.
 
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You started your set off wrong. It's not {(12),(34),...}, it's {(12),(23),...}. Now you need to show any 2 cycle is generated by your set.
 


Oh, my bad. I wrote the set down wrong.

WTS: Any 2-cycle is generated by the set {(12)(23)...(n-1 n)}

So taking any 2-cycle (i j) where j > i (I'm going to work on the difference between j and i. I guess what follows is a sketch. I'll clean it up later.)

If j-i=1, then (i j) = (i i+1). (Probably want to induct on j-i)
Say the above is the base case, then we can assume its true for all (i j) where j-i \leq k (for strong induction), where k \geq1.
There for any (i j) with j-i = k+1, then (i i+1)(i+1 j) and j-(i+1) = j-i-1\leq k.
Then (i+1 j) is a product of 2-cycles of consecutive integers (as in the above set). Then by induction, so is (i j).

Then any (i j) in S_n is a product of consecutive integers. (Fin)
 


Edellaine said:
Oh, my bad. I wrote the set down wrong.

WTS: Any 2-cycle is generated by the set {(12)(23)...(n-1 n)}

So taking any 2-cycle (i j) where j > i (I'm going to work on the difference between j and i. I guess what follows is a sketch. I'll clean it up later.)

If j-i=1, then (i j) = (i i+1). (Probably want to induct on j-i)
Say the above is the base case, then we can assume its true for all (i j) where j-i \leq k (for strong induction), where k \geq1.
There for any (i j) with j-i = k+1, then (i i+1)(i+1 j) and j-(i+1) = j-i-1\leq k.
Then (i+1 j) is a product of 2-cycles of consecutive integers (as in the above set). Then by induction, so is (i j).

Then any (i j) in S_n is a product of consecutive integers. (Fin)

That could use some clean up. I'm having a hard time following it. The basic fact you need is just (j,j+1)(j,k)(j,j+1)=(j+1,k), right?
 
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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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