Question

10.) Consider (Zn; n), the set Zn with mod n multiplication.
i. Argue that if neither a nor b has any common divisors greater than 1
with n then neither does ab. [Equivalently gcd(a; n) = 1, etc.]
ii. Argue that if a does not have any common divisors greater than 1
with n, then [a]n has a multiplicative inverse in Zn.
iii. Argue that (i) and (ii) imply that the set of elements
f[a]n 2 Znjgcd(a; n) = 1g with mod n multiplication is a group.

10.) Consider (Zn, 'n), the set Zn with mod n multiplication. i. Argue that if neither a nor b has any common divisors greater than 1 with n then neither does ab. Equivalently godla, n) = 1, etc.] ii. Argue that if a does not have any common divisors greater than 1 with n, then [a]n has a multiplicative inverse in Zn. iii. Argue that (i) and (ii) imply that the set of elements {[an € Zn|gcd(a, n) = 1} with mod n multiplication is a group.

Answer 1

(10) Consider 'n mod n multiplication 1 the set Zn with Suppose that god ca,n) = 1 = 1 i god (b, n = 1 ged lab, n) =1 Claim : Since gcd laini lain) = 1 .. unio, E Z э aul + nog =1 (2 - Similarly bugt nog Multiplying 10 & 12 we get (ab) lugeug ) n (o, bug tho, ugt buga)=1 Let d be comman any divisor of ab &n Then d I ab & din = d / 1 d = 1 = 1 Hence ged lab, nl

(ii) Suppose ged (and=1 u, u E Then 7 a u the = 1 5 [aut no] = [1 | [au] + [no] [!] [a] [u] + [o ] [a] [u] [1] [1] V Hence, [a] has multiplicative inverse in En liiii Let S = { [a] { En zn I gcd (a, n = 1 S Claimi mod u is multiplictul a with group let -) [a], [b] Es ged (a.nl =1&ged (bin 1 = 1 ged lab, n=1 [ab ES [a].[6] ie ES روس اور امر Hence S is closed under modn multipli- cation

holds ie T Associativity clearly [a] [b] ) [c] [a] ( [b] [c]) Now [IJ E S ged (lino = 1 [a] [ - [a. l] =[a + [a] GS [I] is identity. now, since bu J [b]t Zin let [a] ES ged can l=1 [a] [b [1] Claim i [b]G s le こ gcd Cb, n=1 How, [a] To] =[' =) [ab] : [1] n / ab So, 3 k nk = I- ab =) nk tab=1 Led d= Jed cd (bin) d / b & din d / nkt ab Then =) d/ 1

-) a=1 Hence god (bin) = 1 .. СьЈ E Hence, we con clude s is a dhoe lo