Find an automorphism "Let a and b be two generators of a finite cyclic group G. Show that there exists an automorphism of G that maps a to b"
Im thinking that I need to find a $\phi: G \rightarrow G$, if its possible to show that this $\phi$ is an isomorphism, then I have showed that there exists an automorphism. But how should I think to find such a $\phi$?
Maybe I should note that I know that G is abelian.
 A: Hint: If $G$ is cyclic and generated by $x$ (and we write the group law as addition) then every element of $G$ can be written in the form $nx$ for some $n \in \mathbb Z$.
Similarly if $y$ is some other generator then every element of the group can be written in the form $ny$ for some $n \in \mathbb Z$.
A: As $a$ is a generator, $b = a^n$ for some $n$. Define the map $b^i \mapsto b^{ni}$. This is a homomorphism(Prove).
Similarly define the inverse. Or show directly that the above map is surjective.
A: Before beginning, I'll observe the following. Let $a$ be a generator of a finite cyclic group $G$. Now let $\phi$ be an endomorphism of $G$. Using the fact that $a$ generates $G$, we can characterize the image of any $x = a^r \in G$ like so: $$\phi(x) = \phi(a^r) = \phi(a)^r$$
(you should check that we don't get something different if we represent $x$ with a different power of $a$)
Therefore each endomorphism is exactly determined by specifying $\phi(a)$.
In your context then, just let $\phi$ be the endomorphism determined by $\phi(a) = b$. By construction it maps $a$ to $b$, you just need to check that it's an automorphism.
