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

[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

Homework Equations

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?

 

[fresh_42]

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 Z_n 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

 

[Prof. 27]

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.

 

[fresh_42]

Looks a lot better, thanks.

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 $G/H = \{\,H\, , \,(1,2)H\,\} \cong \mathbb{Z}_2$.