Question

Let T be a set with at least three elements. Show that the permutation group A(T) (Exercise 12) is nonabelian.

Short answer

It is proved that the permutation group is nonabelian.

Step-by-step solution

Condition for abelian and nonabelian group

The group $\mathit{G}$ is said to be abelian if$\mathit{a}\mathit{b}\mathbf{=}\mathit{b}\mathit{a}$ for all $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathbf{G}$, and the group$\mathit{G}$ is said to be nonabelian if$\mathit{a}\mathit{b}\cancel{\mathbf{=}}\mathit{b}\mathit{a}$ for all $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathbf{G}$.

Show that the permutation group is nonabelian

It is known that there are three different elements $a,b,c$in $S$.

Let’s define role="math" localid="1654253152527" $f:T\to T$such that role="math" localid="1654253148909" $f\left(x\right)=\left\{\begin{array}{l}\begin{array}{l}b,x=a\\ c,x=b\\ a,x=c\\ x,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}otherwise}\end{array}\\ \end{array}\right.$and role="math" localid="1654253244491" $g:T\to T$such that $g\left(x\right)=\left\{\begin{array}{l}a,x=a\\ c,x=b\\ b,x=c\\ x,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}otherwise}\end{array}\right.$

Find role="math" localid="1654253291561" $(f\circ g)\left(a\right)$as follows.

$\begin{array}{c}(f\circ g)\left(a\right)=f\left(g\left(a\right)\right)\\ =f\left(a\right)\\ =b\end{array}$

Find$(g\circ f)\left(a\right)$as follows.

$\begin{array}{c}(g\circ f)\left(a\right)=g\left(f\left(a\right)\right)\\ =g\left(b\right)\\ =c\end{array}$

It is observed that $(f\circ g)\left(a\right)\ne (g\circ f)\left(a\right)$.

Thus, it is proved that the permutation group$A\left(T\right)$ is nonabelian.