Group Generators: Showing a^r is Generator of G

[teleport]

Homework Statement

Let G be a finite cyclic group of order n. Let a be a generator. Let r be an integer, not zero, and relatively prime to n.

a)Show that a^r is also a generator of G.

Homework Equations

The Attempt at a Solution

For the first one I need to show that for any element b in G there exists some integer m s.t. (a^r)^m = b. I have no idea what to do. The great thing about this is that I will be tested on this tomorrow. Ha, you get it. Pleast try to answer as soon as possible. Thanks for any help.

 

[Dick]

b=a^s, since a generates G. Can you solve r*m=s mod n, where r is relatively prime to n. Hint: does r have an inverse in Z(n)?

 

[teleport]

Thanks. But we haven't gotten around Z(n) yet. Don't know what that is. Not sure I'm following you either. Does the solution require knowledge of Z(n) or is there any other way around it? Thanks again.

 

[Dick]

You need to know that if r and n are relatively prime, then there is an integer t such that t*r=1 mod n.

 

[teleport]

oh, so now we multiply both sides of the eqn by s:

s*t*r = s mod n

and let m = s*t. Now we get a^(str) = a^(mr) = a^(s mod n) = a^s.

We have shown for any b in G s.t. b = a^s for some s, b = a^rm for some m.

Is that good?

 

[teleport]

Wouldn't we have to show also that the period of a^r is n, to make sure that a^r is not generating something with greater size than G?

 

[teleport]

Oh, because a^r is an element of G, it can only generate subgroups which cannot have greater order than n. ok its clear. Thanks a lot.