Define the set of all affine real-valued functions

$G:=\{f_{a,b} \mid a,b \in \mathbb{R},a \neq 0\}$

where $f_{a,b} : \mathbb{R}\rightarrow \mathbb{R}$ is defined by $f_{ab}: x \mapsto ax+b$.

Is this a group under composition $\circ$?

  • $\begingroup$ What does "under composition 0" mean? And what is the actual question? $\endgroup$
    – user142299
    Apr 11, 2014 at 14:23
  • $\begingroup$ @user142299 I added what I assumed was the question being asked. $\endgroup$
    – Hayden
    Apr 11, 2014 at 14:26
  • $\begingroup$ You only have to check that the group axioms hold $\endgroup$ Apr 11, 2014 at 14:28

1 Answer 1


Well, the identity is $f(x)=x$ because $f(g(x))=g(x)$ and $g(f(x))=g(x)$. The inverse of $$f(x)=ax+b$$ is $$f^{-1}={1\over a}x-{b\over a}$$ which always exists since $a\ne 0$. And function composition is associative.

  • $\begingroup$ Perhaps more clear to denote the identity by $f_{1,0}$, and $f_{a,b}^{-1}$ by $f_{a^{-1},b/a}$. $\endgroup$
    – ah11950
    Apr 11, 2014 at 14:29

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