What are the identity elements in S_3 for x^2=e and y^3=e?

[fk378]

Homework Statement

In S_3, show that there are four elements satisfying x^2=e and three elements satisfying y^3=e.

The Attempt at a Solution

I don't understand what the question is asking at all...

 

[bdforbes]

S_3 is the group of permutations of three symbols, ie {1,2,3}. You can write it out in terms of cyclic permutations:

S_3={(1), (12), (13), (23), (123), (132)}

The question is asking you to show that of those six elements in S_3, four have the property that when squared equal (1), and three have the property that when cubed equal (1).

Before trying to prove anything, play around with the elements a bit, multiply them together, see what you get. Eg:

(1)(12)=(12)
(12)(23)=(123)
(12)(12)=(1)

http://en.wikipedia.org/wiki/Dihedral_group_of_order_6

 

[fk378]

I don't understand the multiplication you did.

(12)(23)=(123)?

Can you explain this a bit further?

 

[bdforbes]

You start with the rightmost permutation ie (23). What does this do to 1? It sends it to 1, since it does not appear in (23). Move to the left. (12) sends 1 to 2. So the product (12)(23) sends 1 to 2. Do this again for 2. (23) sends 2 to 3, (12) doesn't affect 3, to (12)(23) sends 2 to 3. Likewise 3 goes to 1. We write this as (123).

 

[fk378]

Why wouldn't (23) send 2 to 2? Are we working with mod 6, or is this something different?

 

[HallsofIvy]

Where did you get this problem? You seem to be saying that you do not know anything about permutations and have no idea what S₃ is. Did you accidently get your homework from the wrong class?

(23) is the permutation that sends 2 to 3 and 3 to 2 (and sends any other numbers to themselves). No, permutations have nothing whatever to do with "modulo" arithmetic.

 

[bdforbes]

Each cyclic permutation (abc) only acts once in any given product. I echo HallsofIvy's sentiments, you should make sure this question is required in your class.