The Group of Affine Real-Valued Functions on R

# The Group of Affine Real-Valued Functions on R

Let:

(1)\begin{align} \quad G = \{ f_{m, b}(x) = mx + b : m, b \in \mathbb{R}, \: m \neq 0 \} \end{align}

With the operation of function composition, $(G, \circ)$ becomes a group called the **Group of Affine Real-Valued Functions on $\mathbb{R}$**.

Let $f_{m, b}, f_{n, c} \in G$. Then:

(2)\begin{align} \quad (f_{m, b} \circ f_{n, c})(x) = f_{m, b}(f_{n, c}(x)) = f_{m, b}(nx + c) = m(nx + c) +b = mnx + (mc + b) \in G \end{align}

So $G$ is is closed under composition. Since composition of functions is always associative, all that remains to show is that $G$ has an identity and contains inverses of each of its elements.

Note that $f_{1, 0} \in G$, $f_{1, 0}(x) = x$ words as an identity, since for all $f_{m, b} \in G$ we have that:

(3)\begin{align} \quad f_{m, b}(f_{1, 0}(x)) = f_{m, b}(x) = f_{1, 0}(f_{m, b}(x)) \end{align}

Lastly, if $f_{m, b} \in G$ then $f_{1/m, -b/m} \in G$ is such that:

(4)\begin{align} \quad f_{m, b}(f_{1/m, -b/m}(x)) = f_{m, b}((1/m)x + (-b/m)) = m((1/m)x + (-b/m)) + b = m(1/m)x + m(-b/m) + b = x = f_{1, 0}(x) \end{align}

(5)
\begin{align} \quad f_{1/m, -b/m}(f_{m, b}(x)) = f_{1/m, -b/m}(mx + b) = (1/m)(mx + b) + (-b/m) = x = f_{1,0}(x) \end{align}

So indeed, $(G, \circ)$ is a group.