Proving that a finite group is abelian

Question

Having the group $(G,*)$ with $2014$ elements and the endomorphism $f:G \to G,\,f(x)=x^9$. How can you prove that $G$ is abelian?

Answer 1

Hint 1: Note that $9$ and $2014$ are coprime, and that $x^{2014}=e$ for all $x$.

Hint 2:

The above implies that $f$ is an isomorphism, and hence that $f\circ f$ is an isomorphism, and hence that $f\circ f\circ f$, etcetera.

Hint 3:

Find $k$ such that $9^k\equiv7\pmod{2014}$, using the factorization of $2014$.

Answer 2

Let $a \in G$, then $f(a) = a^{9} = e$ if and only if $a = e$

Well, $|a|$ divides $9$ and $|a|$ divides $2014$, but $(9,2014) = 1$ then $|a| = 1$

So $a = e$

Then $f$ is monomorphims between $G$ and $G$, then it is an isomorphims

Let $x,y \in G$.

So $x^{9}y^{9} = f(x)f(y) = f(xy) = (xy)^{9}$

Then $x^{9}y^{9} = (xy)^9$ $ \forall $ $x,y \in G$

And... I can no longer proceed