Exercise 16.1.7. Suppose that p is a prime, and g is a generator of U(p). Consider the function h: U(p) → U(p) given by h(n)= mod (g",p). (Note that h is the same as f defined above, only the domain has been restricted.) Show that h is a bijection. 0
Exercise 16.1.7. Suppose that p is a prime, and g is a generator of U(p). Consider the function h: U(p) → U(p) given by h(n)= mod (g",p). (Note that h is the same as f defined above, only the domain has been restricted.) Show that h is a bijection. 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please do Exercise 16.1.7 and please explain
Transcribed Image Text:This exercise suggests the following question: Under what conditions can
we guarantee that the discrete exponentiation function is onto and/or one-
to-one? (This turns out to be more than just an idle question, as we shall see
shortly.) To gain some leverage against this problem, we will take advantage
of Proposition 20.6.3 from Chapter 20, which tells us that the multiplicative
group Zp\{0} is cyclic, whenever p is a prime. (In Chapter 20 we also used
the notation U (p) instead of Zp\ {0}, and we will use this same notation in
the following.) This means that for any prime p, there is a g EU (p) such that
g is a generator of U(p): that is, U(p) = (g) (recall from Chapter 15 that
for a finite group, (g) = {9, 9², 9³,...}). A generator of U(p) is also referred
to as a primitive root of Zp. Any element of U(p) may be expressed as
a power of g (under mod p multiplication). In other words, the discrete
exponentiation function f : N → U(p) given by f(n) = mod(g", p) is an
onto function!
It turns out that onto-ness also gives use one-to-oneness, when we restrict
f to the appropriate domain:
Exercise 16.1.7. Suppose that p is a prime, and g is a generator of U(p).
Consider the function h: U(p) → U(p) given by h(n) = mod (g", p). (Note
that h is the same as f defined above, only the domain has been restricted.)
Show that h is a bijection.
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