A Permutation Associated With GF \((2^n)\)

If \(\alpha\) is a primitive root of the finite field \({GF}(2^n)\), we define a function \(\pi_n\) on the set \({E}_n = \{1, 2, \ldots, 2^n – 2\}\) by
\[
\pi_\alpha(i) = j \quad \text{iff} \quad \alpha^i = 1 + \alpha^{j}.
\]
Then \(\pi_\alpha\) is a permutation of \({E}_n\) of order \(2\). The path-length of \(\pi\), denoted \({PL}(\pi)\), is the sum of all the quantities \(|\pi(i) – i|\),
and the rank of \(\pi\) is the number of pairs \((i, j)\) with \(i \pi(j)\). We show that \({PL}(\pi) = {2(2^n – 1)(2^{n-1} – 1)}/{3}\), and the rank of \(\pi\) is \((2^{n-1} – 1)^2\).

If \(\gcd(k, 2^n – 1) = 1\), then \(M_k(x) = kx(\mod{2^n – 1})\) is a permutation of \({E}_n\). We show that a necessary condition for the function \(f_i(x) = 1 + x + \cdots + x^{i}\) to be a permutation of \({GF}(2^n)\), is that the function \(g_k(r) = \pi(M_{k+1}(r)) – \pi(r)\) be a permutation of \({E}_n\) such that exactly half the members \(r\) of \({E}_n\) satisfy \(g_k(r) r\).