Question

Let $H=\left\{T_{1,b}|b\in \mathrm{ℝ}\right\}$(notation as in Exercise 33). Prove that H is an abelian group under composition of functions.

Short answer

Itis proved that H is an abelian group under composition of functions.

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 $a,b\in \mathit{G}$.

Show that H is an abelian group under composition of functions.

Check that H satisfies the axioms of a group.

(i)

Let $T_{1,a}$and $T_{1,b}$ be arbitrary elements in H. Then

$\begin{array}{rcl}\left(T_{1,a}◦T_{1,b}\right)\left(x\right)& =& \left(x+b\right)+a\\ & =& x+\left(b+a\right)\in g\end{array}$

(ii)

Associative holds because composition of mapping is associative.

(iii)

Consider $T_{1,0}$.

It is obvious that$f◦T_{1,0}=T_{1,0}◦f=f$ for arbitrary f in H.

(iv)

Let$T_{1,a}$ be arbitrary element in G. So, find its inverse.

$\begin{array}{rcl}{T}_{1,a}◦T_{1,-a}& =& T_{1,-a}◦T_{1,a}\\ & =& T_{1,0}\end{array}$

Find$\left(T_{1,a}◦T_{1,b}\right)\left(x\right)$ as follows.

$\begin{array}{rcl}\left(T_{1,b}◦T_{1,a}\right)\left(x\right)& =& \left(x+b\right)+a\\ & =& \left(x+a\right)+b\end{array}$

It is observed that $\left(T_{1,a}◦T_{1,b}\right)\left(x\right)=\left(T_{1,b}◦T_{1,a}\right)\left(x\right)$.

Thus, it is proved that H is an abelian group under composition of functions.