Proving one element in the symmetric group (s>=3) commutes with all element

In summary, to prove that the only element in Sn (with n>=3) commuting with all the other elements of this group is the identity permutation id, one can either use micromass's solution or show that for any permutation \sigma in Sn other than the identity, there exists a transposition \tau that switches two elements in the disjoint cycle representation of \sigma, thus proving that \sigma does not commute with all other elements.
  • #1
emath
2
0
I am really stuck with how to prove that the only element in Sn (with n>=3) commuting with all the other elements of this group is the identity permutation id.

I have no idea what I am supposed to do with it, i know why S3 has only one element that commutes but i don't know how to prove it for all.
 
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  • #2
You could start by proving the following:

If f and g are in Sn, then the disjoint cycle representation of fgf-1 is obtained by taking the disjoint cycle representation of g and by changing every number n in the representation by g(n).

Think how that would prove your claim.
 
  • #3
I think micromass's solution is very good. If you have trouble with his solution, then you can also just do the following:

If [itex]\sigma \in S_n \setminus \{\mathrm{id}\}[/itex], then there exists an [itex]i \in \{1,\dots,n\}[/itex] such that [itex]\sigma(i) = j[/itex] and [itex]i \neq j[/itex]. Now let [itex]k \in \{1,\dots,n\}[/itex] be such that [itex]k \neq j,\sigma(j)[/itex] and let [itex]\tau \in S_n[/itex] be the transposition which switches [itex]j[/itex] and [itex]k[/itex]. The rest of the proof is trivial from here.
 

1. What does it mean to commute in the symmetric group (s>=3)?

In the symmetric group, elements commute if the order in which they are applied does not affect the final outcome. In other words, if element A and element B commute, then AB = BA.

2. Why is proving commutativity important in the symmetric group?

Proving commutativity in the symmetric group is important because it allows for easier calculations and simplification of group operations. It also provides insight into the structure and properties of the group.

3. How do you prove that one element in the symmetric group commutes with all other elements?

To prove that one element in the symmetric group commutes with all other elements, we must show that the element follows the commutative property when multiplied with any other element in the group. This can be done by using mathematical induction or by showing that the element follows the commutative property for specific cases.

4. Can an element in the symmetric group commute with itself?

Yes, an element in the symmetric group can commute with itself. This is because the commutative property states that an element multiplied with itself will always result in the same outcome, regardless of the order in which the elements are multiplied.

5. Are there any exceptions to proving commutativity in the symmetric group?

Yes, there are exceptions to proving commutativity in the symmetric group. One exception is the identity element, which commutes with all other elements in the group by default. Another exception is when the group is non-abelian, meaning that not all elements in the group commute with each other.

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