3 ms·
But the group has order 2^k. Since the powers of x (for any group element x) form a group themselves, Lagrange's theorme tells us that the order of x must divid
by gizmo686 10y ago
But the group has order 2^k. Since the powers of x (for any group element x) form a group themselves, Lagrange's theorme tells us that the order of x must divide 2^k. This should mean that it is impossible for x to have order (2^k)-1. If fact, it means that x^(2^k)=1.
Further, if you are given the value y=x^2, then no power of y will give you x.
- tromp 10y agoThe finite field GF(2^k) has order (number of elements) 2^k, but its multiplicative group has the (2^k)-1 nonzero elements, and thus order (2^k)-1, which squares [pun intended] with the fact that x^(2^k-1) = 1.
- deleted 10y ago[deleted]