# Generator of the additive gr. Z_m

1. Nov 20, 2008

### sutupidmath

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.

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.

2. Nov 20, 2008

### sutupidmath

Well, i know the previous post is way too long, so if you don't have time to have a look at it, the whole thing is about the following:

Prove that
$$\bar a$$ is a generator of the additive group $$Z_m$$ if and only if gcm(m,a)=1.

3. Nov 20, 2008

### sutupidmath

Nevermind, i figuret it out. Thnx though!