(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

[itex]\mathbb{A} = \{ g_{ab} : (a, b) \in \mathbb{R}^2, \, a \neq 0 \}[/itex]

[itex]g_{ab}(x) = ax + b[/itex]

3. The attempt at a solution

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

Since [itex]f \circ g = g \circ f[/itex], we find that

[itex]f(g(x)) = g(f(x))[/itex]

[itex]f(cx + d) = g(ax + b)[/itex]

[itex]a(cx + d) + b = c(ax + b) + d[/itex]

[itex]acx + ad + b = cax + cb + d[/itex].

This reduces to [itex]ad + b = cb + d[/itex].

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

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# Prove a function is the identity

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