[Math] Prove that a function is bijective and show that G is a group

abstract-algebrafunction-and-relation-compositiongroup-theoryinverse function

Let $a\in\mathbb{R}^∗$ and $b\in \mathbb{R}$. Consider the function $f_{a,b} \in \operatorname{Fun}(\mathbb{R},\mathbb{R})$ given by $f_{a,b}(x)= ax + b$.

a) Show that $f_{a,b}$ is a bijection, and find its inverse function.

b) Let $G$ be the set of functions $\{f_{a,b}|a \in \mathbb{R}^∗
, b \in \mathbb{R}\}$. Show that $G$ is a group, where the group operation is composition of functions. (Thus $G$ is a subgroup of $\operatorname{Bij}(\mathbb{R}, \mathbb{R})$.)

My attempt to solve part a:

Let $x_1, x_2 \in \mathbb{R}$ and assume $x_1 \ne x_2$. Then $f_{a,b}(x_1) = ax_1+b$ and $f_{a,b}(x_2) = ax_2+b$. Thus $f_{a,b}(x_1) \ne f_{a,b}(x_2)$ and hence, $f_{a,b}$ is injective.

Since $a\in \mathbb{R}^∗$ and $b\in \mathbb{R}$, then $ax+b\in \mathbb{R}$ and thus $f_{a,b}(x)=ax + b\in \mathbb{R}$ and since it is a mapping from $\mathbb{R}$ to $\mathbb{R}$ then $f_{a,b}(\mathbb{R})=\mathbb{R}$ which means that $f_{a,b}$ is surjective.

Therefore, $f_{a,b}$ is at a time injective and surjective which means it is bijective.

It's inverse function:
$$f_{a,b}^{-1}(x)=\frac{x-b}{a}$$

Could you please check if I made any mistake in the proof and please give some insights on how to solve part b? thank you!

Best Answer

Your proof of injectivity is ok, but for me it is not that obvious why $f_{a, b} (x_1) \neq f_{a,b} (x_2)$. Your surjectivity proof is wrong. Here is my version:

Injectivity: Let $ax_1 + b = a x_2 + b$. Then $ax_1 = a x_2$ and since $a \neq 0$, it follows $x_1 = x_2$.

Surjectivity: Let $z \in \mathbb{R}$. Choose $x_0 = \frac{z - b}{a}$. Then $f(x_0) = z$. This holds for every $z$, so $f$ is surjective.

So $f$ is bijective. The inverse map is

$$f^{-1} : \mathbb{R} \longrightarrow \mathbb{R},\ x \longmapsto \frac{x - b}{a}.$$

Note that if you show $f \circ f^{-1} = f^{-1} \circ f = \mathrm{id}$ at the beginning, you are already done.

For proving that $G$ is a group, you have to show that

  • $G$ is stable under composition.
  • Composition is associative, i. e., $f \circ (g \circ h) = (f \circ g) \circ h$.
  • In $G$ exists a neutral element $e$ such that $f \circ e = e \circ f = f$.
  • Every $f \in G$ has an inverse element $g \in G$, i. e., $f \circ g = g \circ f = e$. Note that you therefore should use what you have shown before.

I show you the first one and let the rest for you. So let $f_1, f_2 \in G$ with $f_1(x) = ax + b, \ f_2(x) = cx + d$ with $a,b, c, d \in \mathbb{R}$ and $a, c \neq 0$. Then

$$f_1 \circ f_2 (x) = f_1 (f_2(x)) = f_1(cx + d) = a (cx + d) + b = acx + ad + b = \lambda x + \mu \in G, $$

with $\lambda = ac$ and $\mu = ad + b$.

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