An automorphism on a group $G$ is a mapping $\Phi:G\to G$ which is one-to-one, onto and has the property that $\Phi(a\cdot b)=\Phi(a)\cdot\Phi(b)$ for all $a,b \in G$.
Show that if $G$ is Abelian, then the map $\Phi:G\to G$ defined by $\phi(a)=a^{-1}$ for each $a\in G$, is a group automorphism.
I believe this question is wanting me to proof something, but I don't know what I am proving in the second line. Nor do I understand what format they are asking for.
HINT
What are you getting stuck on? You get a definition, and must verify that the function $\phi$ provided fits the definition provided.
UPDATE
A map $G \to G$ is a function that takes elements from $G$ and returning elements of $G$. In our case, certainly, since $(G,*)$ is a group, we know that for each $g \in G$, there exists a unique $g^{-1} \in G$. Thus, the relation $f:G \to G$ given by $\phi(x) = x^{-1}$ is well-defined. To check that it's a function, you need to make sure that $\phi$ maps each $x$ to exactly one element of $G$, and the uniqueness of the inverse guarantees that, so indeed our $\phi$ is a function and is $G \to G$, so the first question is definitely satisfied.