Question

Question 25: Show thatthe subgroup G of${\mathbf{S}}_4$generated by the elements$\mathbf{\sigma }\mathbf{=}\left(\mathbf{1234}\right)$and$\tau \mathbf{=}\left(\mathbf{24}\right)$ has order 8.

Short answer

It is proved that the subgroup G of $S_4$generated by the elements$\sigma =\left(1234\right)$ and$\tau =\left(24\right)$ has order 8.

Step-by-step solution

Step-by-Step Solution Step 1: Statement of theorem 7.23 and theorem 7.25

Theorem 7.23states that when $\sigma =\left(a_1{a}_2\cdots a_k\right)$and $\tau =\left(b_1{b}_2\cdots b_r\right)$can be disjoint cycles in $S_n$, then $\sigma \tau =\tau \sigma$.

Theorem 7.25 states that the order of a permutation$\tau$ in$S_n$ would be theleast common multiple of the lengths of the disjoint cycleswhose product would be${\tau }^{†}$ .

Show that the subgroup G Of S4generated by the elements 0=(12234){"x":[[42,20,10,4,5,15,28,34,31,20],[57,84],[58,84],[116,99,100,114],[267,282,281,265],[124,140,140,140],[147,148,158,172,177,169,157,146,147,176],[186,215,216,186,187,199,210,215,214,202,190,184],[246,244,246,223,223,253]],"y":[[50,49,55,75,103,116,114,82,56,50],[72,72],[94,93],[143,121,39,8],[143,120,37,8],[29,8,9,114],[29,15,8,10,25,50,85,115,115,115],[8,8,8,50,50,49,55,71,107,116,114,100],[116,8,8,81,82,82]],"t":[[0,0,0,0,0,0,0,0,0,0],[0,0],[0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0]],"version":"2.0.0"}and  τ=24has order 8

It should be noted that:

$$\begin{gathered} \left(24\right)\left(1234\right)\left(24\right)={\left(1234\right)}^3 \\ {\left(24\right)}^2=\left(1\right) \\ {\left(1234\right)}^4=\left(1\right) \end{gathered}$$

As a result, the elements in G take the form ${\left(1234\right)}^i{\left(24\right)}^j$for $0⩽i⩽3$and $0⩽j⩽1$. Therefore, there are at most 8 elements in G.It is still necessary to demonstrate that each of these elements is distinct, which could be directly proved as follows:

$$\begin{gathered} {\left(1234\right)}^0=\left(1\right) \\ {\left(1234\right)}^1=\left(1234\right) \\ {\left(1234\right)}^2=\left(13\right)\left(24\right) \\ {\left(1234\right)}^3=\left(1432\right) \\ {\left(1234\right)}^0{\left(24\right)}^1=\left(24\right) \\ {\left(1234\right)}^1{\left(24\right)}^1=\left(12\right)\left(34\right) \\ {\left(1234\right)}^2{\left(24\right)}^1=\left(13\right) \\ {\left(1234\right)}^3{\left(24\right)}^1=\left(14\right)\left(23\right) \end{gathered}$$

Hence, it is proved that the subgroup G of $S_4$ of generated by the elements$\sigma =\left(1234\right)$ and $\tau =\left(24\right)$has order 8