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Let C be the group of continuous bijections on the unit interval to itself using functional composition as the group operation. Let P be the set of all the polynomials in C.

i) Show that P is a group under the operation ii) Show that P is not a subgroup of C iii) Show there are only two first degree polynomials in P and list their formulas iv) Show there are infinitely many quadratic polynomials in P

First of all, if P is closed under the group operation how can it not be a subgroup of C?

For iii) Clearly f(x)=x is one, but I'm lost on the other. I think f(x)=cos(x) or f(x)=sin(x) has promise but it clearly cannot be both...

iv) I don't even know where to begin.

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2 Answers

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To begin on part (iv), write down the formula for the most general quadratic function: $$q(x)=ax^2+bx+c,\qquad a,b,c\in\mathbb{R}.$$ Then ask: what conditions must hold for the coefficients $a,b,c$ in order that $q\in P$? - i.e. in order that

  • $q$ is continuous;
  • $q$ is a bijection of $[0,1]$ onto itself.

Hint: a continuous bijection must be either non-increasing or non-decreasing. Think also about the possible values of $q(0)$ and $q(1)$.

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Either $(i)$ was stated wrong, or you misquoted it. $C$ is a group, not $P$. $P$ is a semigroup.

Hint for $(ii)$: one requirement for a subgroup is that it contains the inverse of any member.

Hint for $(iii)$: if you're thinking of $\sin(x)$ or $\cos(x)$, you're forgetting what "first degree polynomial" means.

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