sutupidmath
Nov20-08, 12:32 AM
Hi all,
I am having trouble proving the following proposition:
\bar a is a generator of the additive group Z_m if and only if gcm(m,a)=1.
Well, first let's start with what i know.
I know how to prove the following:
Let G=[a] be a cyclic group of order q. Show that a^s is a generator of G, iff gcd(s,q)=1.
proof:
(<=) Suppose that gcd(s,q)=1, then there exist some integers x,y such that
1=sx+qy, so
a^1=a^{sx}a^{qy}=(a^s)^x(a^q)^y=(a^s)^xe^y=(a^s)^x so since there exists an integer x, such that (a^s)^x=a then G=[a^s]
(=>) Now, suppose that a^s is a generator of G. THen there should exist some integer k such that
(a^s)^k=a=>a^{sk}=a=>a^{sk-1}=e now from a theorem we know that
q|(sk-1)=> there exists some integer n such that sk-1=mq=>1=sk+(-m)q=>gcd(s,q)=1.
Now i tried to translate this for the additive groups. And here is where the problems start to come in play, for only to get worse when i go to Z_m.
So, i am trying to prove the following:
Let G be an additive group generated by a, where o(G)=q. Prove that s*a is a generator of G iff gcd(s,q)=1.
Proof:
(<=) Suppose that gcd(s,q)=1. Now, as before, there exist x,y integers, such that
1=sx+qy
now: 1*a=(sx+qy)a=(sx)a+(qy)a= x(sa)+y(qa).---(@)
(Now, here i believe that if we Translate the Lagranges theorem into terms of an additive group it would be sth like this, right: "Let G be a finite group with order r. Then the order of each subgroup H in G, and the order of each element a of G is an integral divisor or r. Also r*g=0 for every el. g in G."Basically i am concerned whether the last part would be correct that is: from g^r =e into r*g=e=0. Since in the book we are using we are denoting with 0 the identity in an additive group.)
Now, if this is true, then we get from (@) a=x(sa). Now since there is an integer x, such that this holds, i assume we can conclude that G=[sa], that is sa generates the group G.
(=>) now lets suppose that sa is a generator for the group G. Then there exists some integer k such that k(sa)=a=> k(sa)-a=0 => (ks-1)a=0. Now, since the order of a is q. it follows that
q|(ks-1)=> ks-1=mq =>1=ks +(-m)q , so it follows that gcd(s,q)=1.
Well, let me give a crack to my main issue now:
\bar a is a generator of the additive group Z_m if and only if gcm(m,a)=1.
Proof: again lets suppose that gcd(a,m)=1. so there are integers x, y such that
1=ax+my.
Now, i know that \bar a = a+[m] so let [\bar a]=\{k\bar a:k\in Z\}=\{ka+k(rm):r,k\in Z\} in particular let u\in [\bar a] so, u=ak+k(rm)
So, this would mean that any linear combination of a and m is also in a+[m]. Now since
1=ax+my, i am saying that 1\in a+[m]=\bar a
ok let's stop here, cuz, i lost my stream of thought!
Any hints, ideas, would be greatly appreciated.
I am having trouble proving the following proposition:
\bar a is a generator of the additive group Z_m if and only if gcm(m,a)=1.
Well, first let's start with what i know.
I know how to prove the following:
Let G=[a] be a cyclic group of order q. Show that a^s is a generator of G, iff gcd(s,q)=1.
proof:
(<=) Suppose that gcd(s,q)=1, then there exist some integers x,y such that
1=sx+qy, so
a^1=a^{sx}a^{qy}=(a^s)^x(a^q)^y=(a^s)^xe^y=(a^s)^x so since there exists an integer x, such that (a^s)^x=a then G=[a^s]
(=>) Now, suppose that a^s is a generator of G. THen there should exist some integer k such that
(a^s)^k=a=>a^{sk}=a=>a^{sk-1}=e now from a theorem we know that
q|(sk-1)=> there exists some integer n such that sk-1=mq=>1=sk+(-m)q=>gcd(s,q)=1.
Now i tried to translate this for the additive groups. And here is where the problems start to come in play, for only to get worse when i go to Z_m.
So, i am trying to prove the following:
Let G be an additive group generated by a, where o(G)=q. Prove that s*a is a generator of G iff gcd(s,q)=1.
Proof:
(<=) Suppose that gcd(s,q)=1. Now, as before, there exist x,y integers, such that
1=sx+qy
now: 1*a=(sx+qy)a=(sx)a+(qy)a= x(sa)+y(qa).---(@)
(Now, here i believe that if we Translate the Lagranges theorem into terms of an additive group it would be sth like this, right: "Let G be a finite group with order r. Then the order of each subgroup H in G, and the order of each element a of G is an integral divisor or r. Also r*g=0 for every el. g in G."Basically i am concerned whether the last part would be correct that is: from g^r =e into r*g=e=0. Since in the book we are using we are denoting with 0 the identity in an additive group.)
Now, if this is true, then we get from (@) a=x(sa). Now since there is an integer x, such that this holds, i assume we can conclude that G=[sa], that is sa generates the group G.
(=>) now lets suppose that sa is a generator for the group G. Then there exists some integer k such that k(sa)=a=> k(sa)-a=0 => (ks-1)a=0. Now, since the order of a is q. it follows that
q|(ks-1)=> ks-1=mq =>1=ks +(-m)q , so it follows that gcd(s,q)=1.
Well, let me give a crack to my main issue now:
\bar a is a generator of the additive group Z_m if and only if gcm(m,a)=1.
Proof: again lets suppose that gcd(a,m)=1. so there are integers x, y such that
1=ax+my.
Now, i know that \bar a = a+[m] so let [\bar a]=\{k\bar a:k\in Z\}=\{ka+k(rm):r,k\in Z\} in particular let u\in [\bar a] so, u=ak+k(rm)
So, this would mean that any linear combination of a and m is also in a+[m]. Now since
1=ax+my, i am saying that 1\in a+[m]=\bar a
ok let's stop here, cuz, i lost my stream of thought!
Any hints, ideas, would be greatly appreciated.