Cosets of a subset of S_3

[BustedBreaks]

I am having trouble understanding this example:

Let G=S_3 and H={(1),(13)}. Then the left cosets of H in G are

(1)H=H
(12)H={(12), (12)(13)}={(12),(132)}=(132)HI cannot figure out how to produce this relation:

(12)H={(12), (12)(13)}={(12),(132)}=(132)H

I understand (12)H={(12), (12)(13)} but not how (12)(13) = (132) or the equivalence after that...

 

[daveyinaz]

Note that since (12) and (13) are in S3, they are actually the elements (12)(3) and (13)(2), where 3 and 2 are fixed respectively.

(12)(13), we do computation from right to left, so 1 goes to 3 from (13), then 3 goes to itself in (12). So we have (13...). Now 3 goes to 1 in (13) and 1 goes to 2 in (12), so in (13...) we have 3 goes to 2. Hence (132). This is a quick dirty answer to your question, but I think you should reread or review computations done using permutations.

 

[BustedBreaks]

Note that since (12) and (13) are in S3, they are actually the elements (12)(3) and (13)(2), where 3 and 2 are fixed respectively.

(12)(13), we do computation from right to left, so 1 goes to 3 from (13), then 3 goes to itself in (12). So we have (13...). Now 3 goes to 1 in (13) and 1 goes to 2 in (12), so in (13...) we have 3 goes to 2. Hence (132). This is a quick dirty answer to your question, but I think you should reread or review computations done using permutations.

Okay so I get your method here and I am trying to apply it to this one (23)(13) but I am not getting the answer the book has which is (123)

I set it up like this

(23) (13)
123 123
132 321

then

so 1 goes to 3 then 3 goes to 2, 2 goes to 2 then 2 goes to 3, 3 goes to 1 and 1 goes to 1 so I get 231... I know you said your way was quick and dirty, so maybe I am missing something completely?

EDIT:

Okay so I'm pretty sure that 231 and 123 are the same thing but is there a preference for writing it out?

 

[mrbohn1]

The convention is to start with the smallest number.

 

[blue_raver22]

I would explain this example by first defining what cosets are. Cosets are a way to partition a group into subsets based on a subgroup. In this case, the group G is the symmetric group S_3, which is the group of all permutations of 3 objects. The subgroup H is defined as {(1),(13)}, which means it contains two elements - the identity element (1) and the permutation (13).

Now, to find the left cosets of H in G, we take each element of G and multiply it with each element of H. This results in a set of elements that are all equivalent to each other in terms of the subgroup H. In other words, they all produce the same permutation when multiplied with any element of H.

In the given example, we start with the element (12) in G. When we multiply it with (1) in H, we get (12). When we multiply it with (13) in H, we get (12)(13). So, the left coset of (12) with respect to H is {(12), (12)(13)}. Similarly, we can find the left coset of (132) by multiplying it with (1) and (13) in H, which gives us {(132), (132)(13)}. However, we know that (132)(13) is equivalent to (13)(12) which is equivalent to (12)(13). Hence, we can rewrite the left coset as {(132), (12)(13)}.

Since the left coset of (12) with respect to H is {(12), (12)(13)}, and the left coset of (132) with respect to H is {(132), (12)(13)}, we can conclude that these two left cosets are equivalent. This is because both sets contain the same elements, just in a different order. This is why we can write (12)H={(12), (12)(13)}={(12),(132)}=(132)H.