HW13sol - Homework 13 Math 3360 Applicable Algebra Chapter...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Homework 13: Math 3360 Applicable Algebra Chapter 19: Cyclic Groups and Cryptography 15. Show that the group U 10 of units of Z / 10 Z is cyclic. Find a generator of U 1 0. Solution: Here we have U 10 = { 1 , 3 , 7 , 9 } . Clearly h 1 i = { 1 } , h 3 i = { 3 , 3 2 , 3 3 , 3 4 } = { 3 , 9 , 7 , 1 } , h 7 i = { 7 , 9 , 3 , 1 } , and h 9 h{ 9 , 1 } . Thus we see that h 3 i = h 7 h = U 10 , so U 10 is cyclic and is generated by both 3 and 7. 22. If p is prime, then U p is a cyclic group because there is a primitive root modulo p . Show that U- p has φ ( p- 1) primitive roots. Solution: Let b be a primitive root of U p , so that U p = { b,b 2 ,...,b p- 1 = 1 } . It suffices to show that b k is a primitive root, iff ( k,p- 1) = 1, because φ ( p- 1) is counting the integers smaller than p- 1 which are coprime with it. (Note: showing that b k has order ( p- 1) / ( k,p- 1) is also sufficient, but I didn’t give full credit unless you proved this or cited this result in a theorem). Suppose that ( k,p- 1) = d 6 = 1. Then ( b k ) ( p- 1) /d = ( b p- 1 ) k/d = 1 k/d = 1. Note that this argument works because both ( p- 1) /d and k/d are integers. So we see that b k has order less than p- 1 and hence is not a primitive root....
View Full Document

Page1 / 3

HW13sol - Homework 13 Math 3360 Applicable Algebra Chapter...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online