Chapter 6.2, Problem 6E

a.

To determine

To find the order of $S_4$ , the symmetric group on four elements.

Explanation of Solution

Given:

The symmetric group on four elements.

Calculation:

Let us consider that $S_4=\{1,2,3,4\}$ .

The order of symmetric group of $n$ elements is $n!$ .

Therefore, the order of $S_4$ , the symmetric group on four elements is $4!=24$

b.

To determine

To compute the products $S_4:[1243][4213],[4321][4321],$ and $[2143][1324]$

Explanation of Solution

Let us consider that $f=[1243],g=[4213]$

  $\begin{array}{l}f(1)=1,g(1)=4\\ f(2)=2,g(2)=2\\ f(3)=4,g(3)=1\\ f(4)=3,g(4)=3\end{array}$

Now, finding the product as follows-

  $\begin{array}{l}f\cdot g(1)=f(g(1))=f(4)=3\\ f\cdot g(2)=f(g(2))=f(2)=2\\ f\cdot g(3)=f(g(3))=f(1)=1\\ f\cdot g(4)=f(g(4))=f(3)=4\end{array}$

Therefore,

  $[1243]\cdot [4213]=[3214]$

Similarly,

  $[4321]\cdot [4321]=[1234]$

  $[2143][1324]=[2313]$

c.

To determine

To compute the products $S_4:[3124][3214],[4321][3124],$ and $[1432][1432]$

Explanation of Solution

Let us consider that $f=[3124],g=[3214]$

  $\begin{array}{l}f(1)=3,g(1)=3\\ f(2)=1,g(2)=2\\ f(3)=2,g(3)=1\\ f(4)=4,g(4)=4\end{array}$

Now, finding the product as follows-

  $\begin{array}{l}f\cdot g(1)=f(g(1))=f(3)=2\\ f\cdot g(2)=f(g(2))=f(2)=1\\ f\cdot g(3)=f(g(3))=f(1)=3\\ f\cdot g(4)=f(g(4))=f(4)=4\end{array}$

Therefore,

  $[3124]\cdot [3214]=[2134]$

Similarly,

  $[4321]\cdot [3124]=[2431]$

  $[1432][1432]=[1234]$

d.

To determine

To find the inverses of $[1342],[4123]$ and $[2143]$

Explanation of Solution

The objective is to find $g[abcd]$ such that $f[1342]\cdot g[abcd]=[1234]$

So the inverse of $[1342]$ will be $[1423]$ since $[1342]\cdot [1423]=[1234]$ .

Similarly the inverse of $[4123]$ will be $[2314]$ since $[4123]\cdot [2314]=[1234]$ .

Similarly the inverse of $[2143]$ will be $[2143]$ since $[2143]\cdot [2143]=[1234]$ .

e.

To determine

To show that $S_4$ is not abelian.

Explanation of Solution

For showing $S_4$ not abelian, it needed to be shown as not commutative.

So let us consider

Let us consider that $f=[3124],g=[3214]$

  $\begin{array}{l}f(1)=3,g(1)=3\\ f(2)=1,g(2)=2\\ f(3)=2,g(3)=1\\ f(4)=4,g(4)=4\end{array}$

Now, finding the product as follows-

  $\begin{array}{l}f\cdot g(1)=f(g(1))=f(3)=2\\ g\cdot f(1)=g(f(1))=g(3)=1\end{array}$

Since $f\cdot g(1)\ne g\cdot f(1)$

Hence $S_4$ is not abelian.