S3 Group G/N: Find Left Cosets of N

[namegoeshere]

Homework Statement

Let (G,◦) be a group and let N be a normal subgroup of G. Consider the set of all left cosets of N in G and denote it by G/N:

G/N = {x ◦ N | x ∈ G}.

Find G/N:

(G,◦) = (S₃,◦) and N = <β> with β(1) = 2, β(2) = 3, β(3) = 1.

Homework Equations

The Attempt at a Solution

I'm not sure I understand this problem.

The permutations are (1)(2)(3), (1,2)(3), (1)(2,3), (1,3,2), (1,3)(2), and (1,2,3).

Does β(1) = 2, β(2) = 3, β(3) = 1 mean N is the permutation (1,2,3)? Can't I compose (1,2,3) with any x ∈ G, and receive x ◦ N ∈ G with the resulting quotient group being {x | x ∈ G}, since N is onto and one-to-one?

Sorry if that makes no sense. I'm confused, to say the least.

 

[mighty2000]

Hello,

Thank you for your post! I understand your confusion with this problem. Let me try to explain it in a simpler way.

First, let's define the group (S3,◦) to be the symmetric group on 3 elements, where the operation ◦ represents composition of permutations. This means that (1,2,3) ◦ (1,2) = (1,3) and so on.

Next, let's define the subgroup N to be the set of even permutations, which is generated by the permutation β = (1,2,3). This means that N = {(), (1,2,3), (1,3,2)}. Notice that N is normal in G, since it is a subgroup of the symmetric group (S3,◦).

Now, let's consider the set of all left cosets of N in G, denoted by G/N. This set contains all the possible permutations that can be obtained by composing any permutation in G with the elements of N. In other words, G/N = {(1,2,3) ◦ x | x ∈ G}.

So, the resulting quotient group G/N will contain all the possible permutations that can be obtained by composing (1,2,3) with any permutation in (S3,◦). Since (S3,◦) contains 6 elements, the quotient group G/N will also have 6 elements.

I hope this helps clarify the problem for you. Let me know if you have any further questions. Good luck with your work!