Generating Set for the Symmetric Group - Question

In summary, the permutations (1 2) and (1 2 ... n) generate all of Sn, the symmetric group (the group of all permutations of the numbers {1,2,...,n}), because they can be written as a composition of powers of (1 2 ... n) and (1 2). By conjugating these permutations, all transpositions can be obtained, resulting in all elements of Sn being generated. This pattern applies for all values of n, and the inverse of (1 2 ... n) can also be used in this process.
  • #1
MattL
11
0
Explain why the permutations (1 2) and (1 2 ... n) generate all of Sn, the symmetric group (the group of all permutations of the numbers {1,2,...,n}?

Perhaps something to do with the fact that
(1 2 ... n) = (1 2) (1 3) ... (1 n)?
Other than that I haven't got a clue - help! (please!)

Thanks
 
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  • #2
Try working it out directly for small n; that might give you ideas.
 
  • #3
After giving it a go, I've managed to get, for n = 5

(2 3) = (1 2 3 4 5)^4 (1 2) (1 2 3 4 5)
(3 4) = (1 2 3 4 5)^3 (1 2) (1 2 3 4 5)^2
(4 5) = (1 2 3 4 5)^2 (1 2) (1 2 3 4 5)^3


So I can get the transpositions (1 2), (2 3), (4 5), but I can't work out how to get all of the others that I am missing, these being (1 3), (1 4), (1 5), (2 4), (2 5), (3 5). Even if I could, I'm not sure I'd know how to generalise for n. Also, I'm not entirely sure why the compositions I've used above work or where the pattern comes from - I used my computer to do some guesswork for those.

ps if i don't reply, it's because I've gone to bed!
 
Last edited:
  • #4
Doesn't there seem to be a pattern there? Does it apply for other n? Say, 4 or 6?
 
  • #5
why don't you conjugate (12) by (1,..,n) and powers of (1,..,n)?
 
  • #6
Hurkyl said:
Doesn't there seem to be a pattern there? Does it apply for other n? Say, 4 or 6?

Yep, there seems to be a pattern of
(m+1 m+2) = (1 2 ... n)^(n-m+1) (1 2) (1 2 ... n)^(m-1)
and after trying some in Maple, I think that it does hold for other n.
Still, that doesn't give me a complete set of transpositions.

(Another thing - would be ok to use the inverse of (1 2 ... n) in my answer? Not that I know how I would, but does the original question allow this?)
 
  • #7
The inverse of (1,..,n) is a power of (1,..,n) (but usually one allows inverses as well, though for a finite group this is not explicitly needed - do you see why?).

What makes you think you should get a list of all transpositions from the expressions you've written down? You may then compose (indeed conjugate) the transpositions (m, m+1) and (m+1,m+2) and still be in the set generated by (12) and (1..n)
 
  • #8
matt grime said:
why don't you conjugate (12) by (1,..,n) and powers of (1,..,n)?

how do I conjugate them? I've only recently started permutations and group theory (and yet I've been given this question :confused: ), so my knowledge is still quite basic ( :frown: )

EDIT: oh, I see...
 
  • #9
To conjugate g by h one works out the product h**(-1)gh, where h**(-1) is h inverse. My caret key is broken on my laptop.
 

1. What is a generating set for the symmetric group?

A generating set for the symmetric group is a subset of elements that can be used to generate all elements of the group through repeated application of group operations. In simpler terms, it is a set of elements that can be combined in different ways to create all possible elements of the symmetric group.

2. How is a generating set for the symmetric group different from a normal subgroup?

A normal subgroup is a subset of elements that forms a subgroup and is closed under the group operation, while a generating set is a subset of elements that can be used to create all elements of the group. In other words, a normal subgroup is a subset of the group, while a generating set is a way to describe the entire group.

3. Why is a generating set important in the study of symmetric groups?

A generating set is important because it allows us to describe the entire group using a smaller set of elements. This can make calculations and proofs easier and more efficient. Additionally, studying generating sets can help us understand the structure and properties of the symmetric group.

4. Can a generating set for the symmetric group be finite?

Yes, a generating set for the symmetric group can be finite. In fact, the symmetric group itself is finite, so any generating set for it must also be finite. However, there are also infinite generating sets for the symmetric group, such as the set of all transpositions.

5. How do you find a generating set for a specific symmetric group?

There is no general method for finding a generating set for a specific symmetric group. However, some common techniques include looking for patterns in the elements of the group, using known generating sets for smaller groups, and using properties of the group such as conjugacy or commutativity.

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