Thread: cyclic groups

I have a problem in a textbook that is giving me trouble. Either I'm missing something or there is some sort of mistake in the question. The problem is with part (ii), the other parts seem okay, but I'll post all parts anyway:

Let be a prime number, and let be the group of integers , with multiplication modulo .

(i) Show that if then and deduce that .

(ii) Show that if then and decude that

(iii) Use parts (i) and (ii) to show that the order of the element 2 in is

(iv) Decude that is a power of 2.

Originally Posted by Aradesh

(ii) Show that if then and decude that

Note that thus, note that,
but, is not true

(ii) Show that if then and decude that

it doesn't mean 'then is always true'
it means that if this is true for in , then it implies that . (atleast that is my interpretation of the question)

If p =2^k+1 then 2^k == -1 mod p. If we assume 2^m == 1 mod p with k < m < 2k, then 2^{2k-m} == (-1)^2.(1) == 1 mod p. But 0 < 2k-m < k and by (i) that's impossible. I think there's a sign error in the statement of (ii).

Originally Posted by rgep

If p =2^k+1 then 2^k == -1 mod p. If we assume 2^m == 1 mod p with k < m < 2k, then 2^{2k-m} == (-1)^2.(1) == 1 mod p. But 0 < 2k-m < k and by (i) that's impossible. I think there's a sign error in the statement of (ii).

ie that it should say 1 instead of -1?

Quite so.