# Prove a function is the identity

1. Sep 30, 2011

### tylerc1991

1. The problem statement, all variables and given/known data

Suppose $f$ is one element of $\mathbb{A}$, and it has the property that $f \circ g = g \circ f$ for every $g \in \mathbb{A}$. Prove that $f = e$ (the identity function).

2. Relevant equations

$\mathbb{A} = \{ g_{ab} : (a, b) \in \mathbb{R}^2, \, a \neq 0 \}$
$g_{ab}(x) = ax + b$

3. The attempt at a solution

Choose $f, g \in \mathbb{A}$. By definition, $f$ and $g$ have the form $g_{ab}$ and $g_{cd}$ for some $(a, b) \in \mathbb{R}^2$ and $(c, d) \in \mathbb{R}^2$ such that $a \neq 0$ and $c \neq 0$.

Since $f \circ g = g \circ f$, we find that
$f(g(x)) = g(f(x))$
$f(cx + d) = g(ax + b)$
$a(cx + d) + b = c(ax + b) + d$
$acx + ad + b = cax + cb + d$.

This reduces to $ad + b = cb + d$.

This is where I get stuck. I realize that we are working towards finding $a = 1$ and $b = 0$, so that $g_{ab} = g_{10} = x$ (the identity function). Thank you for your help!

2. Sep 30, 2011

### tylerc1991

I think i figured it out.

If $ad - d = cb - b$ is true for every choice of $c, d$ (as long as $c \neq 0$), then it is true for some. Specifically, we may choose $c = 2$ and $d = 0$ to give $b = 0$. Similarly, we find that $a = 1$.