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Prove a function is the identity

  1. Sep 30, 2011 #1
    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!
     
  2. jcsd
  3. Sep 30, 2011 #2
    I think i figured it out.

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