Question

(a) Show that

$\mathit{V}\mathbf{=}\mathbf{\{}\mathbf{\left(}\begin{array}{cccc}\mathbf{1}& \mathbf{2}& \mathbf{3}& \mathbf{4}\\ \mathbf{1}& \mathbf{2}& \mathbf{3}& \mathbf{4}\end{array}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\begin{array}{cccc}\mathbf{1}& \mathbf{2}& \mathbf{3}& \mathbf{4}\\ \mathbf{2}& \mathbf{1}& \mathbf{4}& \mathbf{3}\end{array}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\begin{array}{cccc}\mathbf{1}& \mathbf{2}& \mathbf{3}& \mathbf{4}\\ \mathbf{3}& \mathbf{4}& \mathbf{1}& \mathbf{2}\end{array}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\begin{array}{cccc}\mathbf{1}& \mathbf{2}& \mathbf{3}& \mathbf{4}\\ \mathbf{4}& \mathbf{3}& \mathbf{2}& \mathbf{1}\end{array}\mathbf{\right)}\mathbf{\}}$is a normal subgroup of ${\mathit{S}}_{\mathbf{4}}$.

Short answer

We proved that$V$ is a normal subgroup of $S_4$.

Step-by-step solution

To show Vis a subgroup ofS4

We have the group .

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

Let us $V$verify is a subgroup of $S_4$.

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

This implies that$V$ is a subgroup of$S_4$

To show Vis a normal subgroup of S4S4 

We have to show that $V$is a normal subgroup of $S_4$.

If $a\in S_4$, then by Exercise 20 of Section 7.4, we have,

$a^{-1}Va$is a subgroup of order 4.

But$V$is the only subgroup of order 4 in $S_4$, where all the other nonidentity elements of$S_4$are of order 3.

Hence, by Corollary 8.6 those elements cannot belongs to the group of order 4.

Then by step 1, we must say that,

.$a^{-1}Va=V,\forall a\in S_4$

Hence, by part (5) of Theorem 8.11, $V$is a normal subgroup of$S_4$ .