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Someone recently asked me how to proof that $x+1$ and $x^3$ generate a free group. A colleague has worked out a proof. I have a vague memory that this has been studied, maybe a Monthly problem? Does anyone know any history on this?

Edit: Sorry for my omission of key point. The group operation here is function composition on the reals, or the integers. Each of the polynomials can be viewed as a permutation of Z = all integers (or on the reals). Viewed in that way, do they generate a free group (of rank two).

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    If your group operation is addition, hopefully it's easy to see that $a(x+1) + b(x^3)$ is never the zero polynomial unless $a=b=0$, so the map $\mathbb{Z}^2 \to (\mathrm{your\,group})$ given by $(a, b)\mapsto a(x+1) + b(x^3)$ is an isomorphism. If you mean under multiplication, then replace $a(x+1) + b(x^3)$ with $(x+1)^a(x^3)^b$ in what I said, and the same still holds. Am I missing something?2012-12-16
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    (Oh - perhaps you mean as *functions* $\mathbb{R}\to \mathbb{R}$ or something, rather than *polynomials*? That sounds less easy!)2012-12-16
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    Minor nitpickery: $x\to f(x)=x^3$ is hardly a permutation on $\mathbb{Z}$, and it certainly doesn't have an inverse on that domain (what is $f^{-1}(2)$?). Either you need to be working over $\mathbb{R}$ or you're considering the free _monoid_, which is a different matter entirely.2013-03-28

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