Question

In Exercises 7-1 $\mathit{G}$is a group and$\mathit{H}$ is a subgroup of G. Find the index $\mathbf{[}\mathbf{G}\mathbf{:}\mathbf{H}\mathbf{]}$.

11.$\mathit{H}$ is the subgroup generated by$\left(\begin{array}{cccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 3\end{array}& \begin{array}{l}3\\ 4\end{array}& \begin{array}{l}4\\ 1\end{array}\end{array}\right)\mathbf{}\mathbf{;}\mathit{G}\mathbf{=}{\mathit{S}}_{\mathbf{4}}$

Short answer

$k$has 6 distinct co-sets

Step-by-step solution

Determine given functions

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

Now let’s consider$S_4$ is the symmetric group of 4 elements.$k$ is the cyclic group and its generated by,$\left(\begin{array}{cccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 3\end{array}& \begin{array}{l}3\\ 4\end{array}& \begin{array}{l}4\\ 1\end{array}\end{array}\right)$

Determine the index

Then,

$$\begin{gathered} \left(\begin{array}{cccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 3\end{array}& \begin{array}{l}3\\ 4\end{array}& \begin{array}{l}4\\ 1\end{array}\end{array}\right)\left(\begin{array}{cccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 3\end{array}& \begin{array}{l}3\\ 4\end{array}& \begin{array}{l}4\\ 1\end{array}\end{array}\right)=\left(\begin{array}{cccc}\begin{array}{l}1\\ 3\end{array}& \begin{array}{l}2\\ 4\end{array}& \begin{array}{l}3\\ 1\end{array}& \begin{array}{l}4\\ 2\end{array}\end{array}\right) \\ \left(\begin{array}{cccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 3\end{array}& \begin{array}{l}3\\ 4\end{array}& \begin{array}{l}4\\ 1\end{array}\end{array}\right)\left(\begin{array}{cccc}\begin{array}{l}1\\ 3\end{array}& \begin{array}{l}2\\ 4\end{array}& \begin{array}{l}3\\ 1\end{array}& \begin{array}{l}4\\ 2\end{array}\end{array}\right)=\left(\begin{array}{cccc}\begin{array}{l}1\\ 4\end{array}& \begin{array}{l}2\\ 1\end{array}& \begin{array}{l}3\\ 2\end{array}& \begin{array}{l}4\\ 3\end{array}\end{array}\right) \\ \left(\begin{array}{cccc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}2\\ 3\end{array}& \begin{array}{l}3\\ 4\end{array}& \begin{array}{l}4\\ 1\end{array}\end{array}\right)\left(\begin{array}{cccc}\begin{array}{l}1\\ 4\end{array}& \begin{array}{l}2\\ 1\end{array}& \begin{array}{l}3\\ 2\end{array}& \begin{array}{l}4\\ 3\end{array}\end{array}\right)=\left(\begin{array}{cccc}\begin{array}{l}1\\ 1\end{array}& \begin{array}{l}2\\ 2\end{array}& \begin{array}{l}3\\ 3\end{array}& \begin{array}{l}4\\ 4\end{array}\end{array}\right) \end{gathered}$$

As order of$S_4$ is$4!=24$ and order of the subgroup$k$ is 4.

By applying Lagrange’s theorem,

And from this$k$ has 6 co-sets.