Question

$\mathit{G}$is a group and$\mathit{k}$ is a subgroup of $\mathit{G}$. List the distinct right co-sets of$\mathit{k}$ in$\mathit{G}$ .

4. $\mathbf{k}\mathbf{=}\left\{\left(\begin{array}{ccc}\begin{array}{l}1\\ 1\end{array}& \begin{array}{l}2\\ 2\end{array}& \begin{array}{l}3\\ 3\end{array}\end{array}\right),\left(\begin{array}{ccc}\begin{array}{l}1\\ 1\end{array}& \begin{array}{l}2\\ 3\end{array}& \begin{array}{l}3\\ 2\end{array}\end{array}\right)\right\}\mathbf{;}\mathbf{G}\mathbf{=}{\mathbf{S}}_{\mathbf{3}}$

Short answer

The list of distinct right co-sets is:

$ke=k$

$$\begin{gathered} k\left(\begin{array}{ccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 1\end{array}& \begin{array}{l}3\\ 3\end{array}\end{array}\right)=\left\{\left(\begin{array}{ccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 3\end{array}& \begin{array}{l}3\\ 1\end{array}\end{array}\right),\left(\begin{array}{ccc}\begin{array}{l}1\\ 3\end{array}& \begin{array}{l}2\\ 1\end{array}& \begin{array}{l}3\\ 2\end{array}\end{array}\right)\right\} \\ k\left(\begin{array}{ccc}\begin{array}{l}1\\ 3\end{array}& \begin{array}{l}2\\ 1\end{array}& \begin{array}{l}3\\ 2\end{array}\end{array}\right)=\left\{\left(\begin{array}{ccc}\begin{array}{l}1\\ 3\end{array}& \begin{array}{l}2\\ 1\end{array}& \begin{array}{l}3\\ 2\end{array}\end{array}\right),\left(\begin{array}{ccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 3\end{array}& \begin{array}{l}3\\ 1\end{array}\end{array}\right)\right\} \end{gathered}$$

Step-by-step solution

Determine given functions

Considering the given functions,$G$is the group and $k$is the subgroup.

Now let’s $S_3$is the symmetric group of 3 elements.

$k=\left\{\left(\begin{array}{ccc}\begin{array}{l}1\\ 1\end{array}& \begin{array}{l}2\\ 2\end{array}& \begin{array}{l}3\\ 3\end{array}\end{array}\right),\left(\begin{array}{ccc}\begin{array}{l}1\\ 1\end{array}& \begin{array}{l}2\\ 2\end{array}& \begin{array}{l}3\\ 3\end{array}\end{array}\right)\right\}$

Determine list of co-sets

As order of $S_3$is $3!=6$and order of the subgroup $k$is 2.

By applying Lagrange’s theorem,

And from this $k$has 3 co-sets, and these co-sets are,

$ke=k$

$$\begin{gathered} k\left(\begin{array}{ccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 1\end{array}& \begin{array}{l}3\\ 3\end{array}\end{array}\right)=\left\{\left(\begin{array}{ccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 3\end{array}& \begin{array}{l}3\\ 1\end{array}\end{array}\right),\left(\begin{array}{ccc}\begin{array}{l}1\\ 3\end{array}& \begin{array}{l}2\\ 1\end{array}& \begin{array}{l}3\\ 2\end{array}\end{array}\right)\right\} \\ k\left(\begin{array}{ccc}\begin{array}{l}1\\ 3\end{array}& \begin{array}{l}2\\ 1\end{array}& \begin{array}{l}3\\ 2\end{array}\end{array}\right)=\left\{\left(\begin{array}{ccc}\begin{array}{l}1\\ 3\end{array}& \begin{array}{l}2\\ 1\end{array}& \begin{array}{l}3\\ 2\end{array}\end{array}\right),\left(\begin{array}{ccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 3\end{array}& \begin{array}{l}3\\ 1\end{array}\end{array}\right)\right\} \end{gathered}$$

Hence, the right distinct co-sets are $ke=k$,$k\left(\begin{array}{ccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 1\end{array}& \begin{array}{l}3\\ 3\end{array}\end{array}\right)=\left\{\left(\begin{array}{ccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 3\end{array}& \begin{array}{l}3\\ 1\end{array}\end{array}\right),\left(\begin{array}{ccc}\begin{array}{l}1\\ 3\end{array}& \begin{array}{l}2\\ 1\end{array}& \begin{array}{l}3\\ 2\end{array}\end{array}\right)\right\}$ and $k\left(\begin{array}{ccc}\begin{array}{l}1\\ 3\end{array}& \begin{array}{l}2\\ 1\end{array}& \begin{array}{l}3\\ 2\end{array}\end{array}\right)=\left\{\left(\begin{array}{ccc}\begin{array}{l}1\\ 3\end{array}& \begin{array}{l}2\\ 1\end{array}& \begin{array}{l}3\\ 2\end{array}\end{array}\right),\left(\begin{array}{ccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 3\end{array}& \begin{array}{l}3\\ 1\end{array}\end{array}\right)\right\}$.