Question

If$a,b\in \mathrm{ℝ}$ with $a\ne 0$, let$T_{a,b}:\mathrm{ℝ}\to \mathrm{ℝ}$ be the function given by $T_{a,b}\left(x\right)=ax+b$. Prove that the setrole="math" localid="1653388144540" $G=\left\{T_{a,b}\right|a,b\in \mathrm{ℝ}$with$a\ne 0\}$ forms a nonabelian group under composition of functions.

Short answer

It is proved that is nonabelian group under composition of function.

Step-by-step solution

Condition for abelian and nonabelian group

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

Show that that the set G={Ta,b|a,b∈ℝ with a≠0} forms a nonabelian group under composition of functions.

Assume that G is group.

Let $T_{a,b},T_{c,d}\in G$. This implies that

$$\begin{gathered} T_{a,b}\left(x\right)=ax+b \\ {T}_{c,d}\left(x\right)=cx+d \end{gathered}$$

Find $\left(T_{a,b}◦T_{c,d}\right)\left(x\right)$as follows.

$\begin{array}{rcl}\left(T_{a,b}◦T_{c,d}\right)\left(x\right)& =& T_{a,b}\left(T_{c,d}\left(x\right)\right)\\ & =& T_{a,b}\left(cx+d\right)\\ & =& a\left(cx+d\right)+b\end{array}$

Find $\left(T_{c,d}◦T_{a,b}\right)\left(x\right)$as follows.

$\begin{array}{rcl}\left(T_{c,d}◦T_{a,b}\right)\left(x\right)& =& T_{c,d}\left(T_{a,b}\left(x\right)\right)\\ & =& T_{c,d}\left(ax+b\right)\\ & =& c\left(ax+b\right)+d\end{array}$

$\begin{array}{rcl}\left(T_{c,d}◦T_{a,b}\right)\left(x\right)& =& T_{c,d}\left(T_{a,b}\left(x\right)\right)\\ & =& T_{c,d}\left(ax+b\right)\\ & =& c\left(ax+b\right)+d\end{array}$

It is observed that $\left(T_{a,b}◦T_{c,d}\right)\left(x\right)\ne \left(T_{c,d}◦T_{a,b}\right)\left(x\right)$.

Thus, it is proved that G is nonabelian group under composition of function.