# cyclic groups

• Jun 11th 2006, 10:13 AM
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 $p = 2^k +1$ be a prime number, and let $G$ be the group of integers $1, 2, \ldots, p-1$, with multiplication modulo $p$.

(i) Show that if $0 < m < k$ then $0 < 2^m -1 < p$ and deduce that $2^m \neq 1 \mod p$.

(ii) Show that if $k< m < 2k$ then $2^m = 1 \mod p \Longrightarrow 2^{2k-m} = -1 \mod p$ and decude that $2^m \neq 1 \mod p$

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

(iv) Decude that $k$ is a power of 2.
• Jun 11th 2006, 10:38 AM
ThePerfectHacker
Quote:

(ii) Show that if $k< m < 2k$ then $2^m = 1 \mod p \Longrightarrow 2^{2k-m} = -1 \mod p$ and decude that $2^m \neq 1 \mod p$

Note that $5=2^2+1$ thus, $k=2$ note that,
$2=k but, $2^3=1 \mbox{ mod } 5$ is not true :eek:
• Jun 11th 2006, 10:57 AM
Quote:

(ii) Show that if $k< m < 2k$ then $2^m = 1 \mod p \Longrightarrow 2^{2k-m} = -1 \mod p$ and decude that $2^m \neq 1 \mod p$
it doesn't mean 'then $2^m = 1 \mod p$ is always true'
it means that if this is true for $m$ in $k, then it implies that $2^{2k-m} = -1 \mod p$. (atleast that is my interpretation of the question)
• Jun 11th 2006, 11:27 AM
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).
• Jun 11th 2006, 12:02 PM