# Isomorphisms between cyclic groups!

1. Nov 16, 2008

### sutupidmath

Ok, here is something i thought i understood, but it turns out i am having difficulties fully grasping/proving it.

Let $$\theta:G->G'$$ be an isomorphism between G and G', where o(G)=m=o(G'), and both G and G' are cyclic, i.e. G=[a] and G'=

So my question is, when we want to find the total number of isomorphisms from G to G', i 'know' that the total number of such isomorphisms is basically
the number of generators of G':
$$\theta(a)=b^k$$ where gcd(k,m)=1. But i don't really understand why? That is, how are we sure that by just counting the number of generators on G' we are actually finding the total number of such isomorphisms. ?

Any explanations would be appreciated.

2. Nov 17, 2008

### HallsofIvy

Staff Emeritus
Suppose $\theta$ were any isomorphism from G to G'. Apply $\theta$ to a generator of G. It is simple to prove that $\theta$ MUST map a generator to a generator. So it is easy to see that it is a matter of how many generators each has: each mapping of generator to generator gives an isomorphism.

3. Nov 17, 2008

### sutupidmath

well, yeah, i know how to prove that an isomorphism perserves the order of each element, and thus must mapp a generator to a generator, but say for example that the generators of G=[a] are a, a^3,a^5, a^7, and the generators of G'= are b, b^3, b^5,b^7, then to me it looks more logical to have these isomorphisms:
first $$\theta(a)=b^i; \theta(a^3)=b^i, \theta(a^5)=b^i;\theta(a^7)=b^i,i=1,3,5,7$$ , which means not 4, but 4! isomorphisms.

So, what am i missing here?

4. Nov 17, 2008

### d_leet

Once you map a generator somewhere you fix where every other element in the group goes. You can't map say a to b and a^3 to b^5 with one homomorphism because the first implies that a^3 must map to b^3.

5. Nov 19, 2008

### sutupidmath

well, yeah, i think i get it.

thnx