Answer to Question #218160 in Abstract Algebra for PRIYANKA

Question

Let G=fg:R--li:g(x)=ax+b,a,b EQ,a# 01. Check whether or not G is a group with respect to the composition of mappings. For f(x) = 2x + 3, find all g bilong toG such that fog=gof

Answer 1

Let $f,g,h\in G$ where

$$f(x)=a_1x+b_1,g(x)=a_2x+b_2, \\h(x)=a_3x+b_3,a_1,b_1,a_2,b_2,a_3,b_3\in \mathbb{R}$$

Consider

$$f\circ g(x)=a_1(a_2x+b_2)+b_1=a_1a_2x+a_1b_2+b_1$$

Clearly, $a_1a_2,a_1b_2+b_1\in \mathbb{R}$ since the sum of product of real numbers is also a real number. Hence $f\circ g\in G$.

Next,

$$g\circ h(x)=a_2(a_3x+b_3)+b_2=a_2a_3x+a_2b_3+b_2$$

So consider

$$f\circ(g\circ h)(x)=a_1(a_2a_3x+a_2b_3+b_2)+b_1 \\=a_1a_2a_3x+a_1a_2b_3+a_1b_2+b_1 \\=a_1a_2(a_3x+b_3)+a_1b_2+b_1 \\=(f\circ g)\circ h(x)$$

Next, let $e\in G$ such that $e(x)=x$. Then we have

$$g\circ e(x)=a_2(e(x))+b=a_2x+b_2=g(x)=a_2x+b_2=e(g(x)) \\=e\circ g(x)$$

Finally, let $k(x)=\dfrac{x}{a_2}-\dfrac{b_2}{a_2}$ where $a\neq 0$. Clearly, $k\in G$. Now consider

$$g\circ k(x)=a_2\left(\dfrac{x}{a_2}-\dfrac{b_2}{a_2}\right)+b_2=x-b_2+b_2=x \\=e(x)$$

Similarly,

$$k\circ g(x)=\dfrac{a_2x+b_2}{a_2}-\dfrac{b_2}{a_2}=x=e(x)$$

Since $G$ satisfies all the conditions of a group, we therefore conclude that $G$ is a group.

Let $g(x)=ax+b$ where $a,b\in \mathbb{R}$. Suppose $f\circ g=g\circ f$. Then we have

$$2(g(x))+3=g(2x+3) \\ \Rightarrow 2(ax+b)+3=a(2x+3)+b \\ \Rightarrow 2ax+2b+3=2ax+3a+ab$$

By comparing the terms of both sides of the equation above, we have

$$2b+3=3a+ab \Rightarrow a(b+3)=2b+3 \\ \Rightarrow a=\dfrac{2b+3}{b+3}$$

where $b\neq -3$.

Hence we have the set

$\{f \in G | f(x)=ax+b \text{ where }a=\dfrac{2b+3}{b+3},b \neq -3\}$