# Finding Cosets of subgroup <(3,2,1)> of G = S3

• Prof. 27
In summary, a coset in group theory is a subset of a group created by multiplying all elements in the subset by a single element in the group. To find cosets of a subgroup in a group, you can multiply each element of the subgroup by each element of the group and combine any duplicate results. The subgroup <(3,2,1)> of G = S3 is a subgroup of the symmetric group S3 that contains all the elements obtained by permuting the numbers 1, 2, and 3. There are 3 cosets of this subgroup in G = S3. Finding cosets in group theory helps to understand the structure of a group and the relationship between subgroups and the larger group, and
Prof. 27

## Homework Statement

Find all cosets of the subgroup H in the group G given below. What is the index (G : H)?
H = <(3,2,1)>, G = S3

## The Attempt at a Solution

I will leave out the initial (1,2,3) part of the permutation. We have S3 = {(1,2,3),(2,1,3),(3,2,1),(3,1,2),(2,3,1),(1,3,2)}

And for H we have

(3,2,1)
(3,2,1)+(3,2,1) = (3,1,2)
(3,2,1)+(3,2,1)+(3,2,1) = (3,3,3)
So H = {(3,2,1),(3,1,2),(3,3,3)}

The problem is that (3,3,3) is not in S3. If I ignore this then I find the cosets:

0 + <(3,2,1)> = {(3,2,1),(3,1,2),(3,3,3)}
1 + <(3,2,1)> = {(1,3,2),(1,2,3),(1,1,1)}
2+ <(3,2,1)> = {(2,1,3),(2,3,1),(2,2,2)}

This exhausts S3 but there are these additional elements not in it. I can't figure out what I'm missing. Any pointers?

Your notation is difficult to read. Do you write a permutation ##\pi## as ##(\pi(1),\pi(2),\pi(3))\,##? And what do you mean by ##0,1,2\,?##

Usually we write a permutation as e.g. ##(1,3)=\begin{pmatrix}1&2&3\\3&2&1\end{pmatrix}\,:\,1 \mapsto 3 \mapsto 1\; , \; 2## is fixed.
So especially what is ##H##? Is it ##\langle (3,2,1) \rangle##, that is ##3 \mapsto 2 \mapsto 1 \mapsto 3## or ##\langle (1,3) \rangle##, which is ##1 \mapsto 3 \mapsto 1\;?##

Prof. 27
Okay, so this makes more sense now. I was in the mindset of Zn when I saw H.
So we have S3 = {(1)(2)(3), (1)(2, 3), (3)(1, 2), (1, 2, 3), (1, 3, 2), (2)(1, 3)}

With H = <(3,2,1)>
Using permutation multiplication to generate H we get:
(3,2,1)^1 = (3,2,1) = (1,3,2)
(3,2,1)^2 = (1,2,3)
(3,2,1)^3 = (1)(2)(3) = id
(3,2,1)^4 = (1,3,2)
(3,2,1)^5 = (1,2,3)
(3,2,1)^6 = (1)(2)(3) = id

H = {(1,3,2), (1,2,3), (1)(2)(3)}

So I need to find some permutations g, p such that
pH = H, so p = (1)(2)(3) = id
And
gH = {(1)(2,3), (3)(1,2), (2)(1,3)}

These will be the two left cosets of H in S3

Is that a correct formulation? Is there a general method you'd use to find this or is it just trial and error?

Thanks so much for the help

Wait I think I just figured it out. Since one of the elements of H is the identity element g must be one of the three elements listed in gH.

Looks a lot better, thanks.
Prof. 27 said:
Since one of the elements of H is the identity element g must be one of the three elements listed in gH.
Yes. In general we have ##G = g_1H \cup g_2H \cup \ldots##, i.e. the set ##G## will be written as disjoint union of subsets ##g_iH##. The ##g_iH## all have the same number of elements. In our case ##|G|=|S_3|=6## and ##|H|=3##, since as you correctly said ##H=\{\,1\, , \,(1,2,3)\, , \,(1,3,2)\,\}##. That leaves us with ##6:3=2=|G/H|=[G\, : \,H]## many subsets ##g_iH##.

Now one of them is clearly ##H=1\cdot H## itself, as ##1 \in G##. The other one has to be ##g\cdot H## with ##g\notin H##, say ##g=(1,2)\,.##
Thus we get as the other coset ##(1,2)\cdot H = \{\,(1,2)(1,2,3)=(2,3)\, , \,(1,2)(1,3,2)=(1,3)\, , \,(1,2) \cdot 1=(1,2)\,\}\,.##

This construction works well with every subgroup of ##G##. Now in case ##H## is a normal subgroup, we can even define a group structure on the set of cosets, that is, ##G/H=\{\,H\, , \,g_1H\, , \,g_2H\, , \,\ldots \,\}## is a again a group with ##g_i H\cdot g_j H =(g_ig_j) \cdot H## . This is the basic difference between a subgroup and a normal subgroup. In our case ##H \cong \mathbb{Z}_3 \triangleleft S_3 =G## is a normal subgroup and ##G/H = \{\,H\, , \,(1,2)H\,\} \cong \mathbb{Z}_2##.

Prof. 27

## 1. What is a coset in group theory?

A coset is a subset of a group that is created by multiplying all elements in the subset by a single element in the group. It is denoted as gH, where g is an element in the group G and H is a subgroup of G.

## 2. How do you find cosets of a subgroup in a group?

To find cosets of a subgroup H in a group G, you can multiply each element of H by each element of G and list the results. Then, you can combine any duplicate results to create the cosets. This process is also known as the left coset multiplication.

## 3. What is the subgroup <(3,2,1)> of G = S3?

The subgroup <(3,2,1)> of G = S3 is a subgroup of G that contains all the elements of G that can be obtained by permuting the numbers 1, 2, and 3. It is a subgroup of the symmetric group S3, which consists of all possible permutations of 3 objects.

## 4. How many cosets are there of subgroup <(3,2,1)> in G = S3?

There are 3 cosets of subgroup <(3,2,1)> in G = S3, as there are 3 elements in G and the subgroup <(3,2,1)> contains all the elements of G.

## 5. What is the significance of finding cosets in group theory?

Finding cosets in group theory helps to better understand the structure of a group by identifying all the possible combinations of elements in the group. It also allows for the study of the relationship between subgroups and the larger group, and can be used to prove important theorems in group theory.

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