Homomorphic Group and the Image of Generator

[gbean]

Homework Statement

Let phi be a homomorphism from the group G under * to the group G' under #, where G = <a>, the cyclic group generated by a. Show that phi is completely determined by the image of the generator a of G.

Homework Equations

Phi is a homomorphism, therefore phi(x*y) = phi(x)#phi(y), or it preserves group structure.
Image: subset of all outputs in the codomain that are mapped to from elements of the domain

The Attempt at a Solution

I'm not sure what the image of a generator is, so I'm stuck on how to start this problem.

 

[micromass]

If G=<a>, then a is called the generator of the group. The problem states that phi is completely determined by the value $$\varphi(a)$$. That is, if you know what $$\varphi(a)$$ is, then you know every other value of phi...

 

[gbean]

If G=<a>, then a is called the generator of the group. The problem states that phi is completely determined by the value $$\varphi(a)$$. That is, if you know what $$\varphi(a)$$ is, then you know every other value of phi...

If G=<a>, then for all b in G, b = a^k.

G' = phi(G)
=> a' = phi(a)
=> phi(a^k) = phi(a)^k
=> G(<a>) = <phi(a)>

Is that all I need to say?

 

[micromass]

Yes, that's it!