Prove that the homomorphism $f:G\to G$, $G$ cyclic group, is an automorphism iff the image of the generatory element is a generatory element.

Question

I have proven that when the homomorphism $f$ is a automorphism, then if $a$ generates the cyclic group $G$, that $f(a)$ also generates the cyclic group G.

The problem is the other direction, if $a$ and $f(a)$ are generators of the cyclic group $G$, then the homomorphism $f$ is an automorphims. I believe that I have to prove that $f$ is a bijective function, but do I define some function or do I approach this differently.

Answer 1

Hint for showing surjectivity: $f(a)^k = f(a^k).$ Hint for showing injectivity: for any elements in the codomain, they are of the form $f(a^m) $ and $f(a^n).$ Multiply one by the inverse of the other and conclude.

Answer 2

Notice that $f(g^n)=f(g)^n$. So if $f(g)$ is a generator then clearly $f$ is surjective.

If $G$ is finite this clearly implies that $f$ is also injective.

If $G$ is infinite notice that $f(g)^n=f(g)^m$ if and only if $n=m$ (otherwise the genrator has finite order). Therefore, if $f(g^n)=f(g^m)$ we have $n=m$, and so $f$ is also injective.