Prove a function is the identity

[tylerc1991]

Homework Statement

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).

Homework Equations

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

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!

 

[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.