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

Click For Summary
SUMMARY

The symmetric group S_n is generated by the set of transpositions {(1 2), (3 4), ..., (n-1 n)}. Any element in S_n can be expressed as a product of disjoint n-cycles, which can further be decomposed into 2-cycles. The proof utilizes strong induction to demonstrate that any 2-cycle (i j) can be represented as a product of consecutive transpositions from the specified set, confirming that S_n is indeed generated by these transpositions.

PREREQUISITES
  • Understanding of symmetric groups and their properties
  • Familiarity with cycle notation in permutations
  • Knowledge of mathematical induction, particularly strong induction
  • Basic concepts of group theory
NEXT STEPS
  • Study the properties of symmetric groups, specifically S_n and their generators
  • Learn about cycle decomposition in permutations
  • Explore the concept of transpositions and their role in group generation
  • Review strong induction techniques and their applications in proofs
USEFUL FOR

Mathematicians, students of abstract algebra, and anyone studying group theory, particularly those interested in the properties and generation of symmetric groups.

Edellaine
Messages
11
Reaction score
0

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.
 
Physics news on Phys.org


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?
 
Last edited:

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
5K
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
7K