# Permutations Sets in Binary Finite Fields

by user0410   Last Updated September 11, 2019 13:20 PM

Let $$a \in \mathbb{F}_{2^n}$$ such that the multiplicative order of $$a$$ is r. Now consider the following two sets:

$${\bf A}=\{0,1,a,a^2,\cdots ,a^{r-1} \}, \quad {\bf B}=\{0,1,a+1,a^2+1,\cdots ,a^{r-1}+1 \}.$$

My question: How to prove that the set $$\bf B$$ is a permutation of the set $$\bf A$$?

Thanks for any help

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This is false in general. Let $$n=4$$, $$r=5$$. If $$\gamma$$ is a zero of the polynomial $$x^4+x+1$$, then $$a=\gamma^3$$ is of order $$r$$. According to a table of the elements of $$\Bbb{F}_{16}$$ we then have $$A=\{0,1,\gamma^3,\gamma^6=\gamma^3+\gamma^2,\gamma^9=\gamma^3+\gamma,\gamma^{12}=\gamma^3+\gamma^2+\gamma+1\}.$$ The presentation of an element of $$\Bbb{F}_{16}$$ as a polynomial in $$\gamma$$ of degree $$<4$$ is unique, so we immediately see that $$B$$ is a different set. For example $$\gamma^3+1\in B, \gamma^3+1=\gamma^{14}\notin A$$.