$H$ is a finite abelian of order $k$, also $q \in Z^{+}$ with $(k,q)=1$. $\phi:H \rightarrow H$ by $\phi(h)=h^q,\forall h \in H$ belongs to $Aut(G)$. $H$ is a finite abelian group of order $k$, also $q \in Z^{+}$ with $(k,q)=1$. $\phi:H \rightarrow H$ by $\phi(h)=h^q,\forall h \in H$ belongs to $Aut(G)$.
What I did so far:
$\phi(h_{1}h_{2})=(h_{1}h_{2})^q=h_{1}^qh_{2}^q=\phi(h_{1})\phi(h_{2})$
If $\phi(h_{1})=\phi(h_{2})$, then $h_{1}^q=h_{2}^q$,therefore $h_{1}=h_{2}$(one to one)
$\forall h\in H$, we have $\phi(h)=h^q$ (onto)
I think I did something wrong here, since there are two conditions that I do not use:1. abelian group; 2. order $k$ and $(k,q)=1$
Does anyone could help me? Thanks!

The condition that $H$ is abelian is used in proving that $H$ is a homomorphism.
Since $(k,q) = 1$, $\exists x,y \in \Bbb Z$ such that $qx + ky = 1$.
If $\phi(h_{1})=\phi(h_{2})$, then $h_{1}^q=h_{2}^q$.
Since $|H| = k$, $h_1^k = h_2^k = e$.
$$h_1 = h_1^{qx+ky}=h_1^{qx}=h_2^{qx} = h_2^{qx+ky}=h_2.$$
Similarly, surjectivity can be proved.
$$\phi(h^x) = h^{qx} = h^{qx + ky} = h$$
Hence $\phi:H\to H$ is an automorphism on $H$.
