Let $f:G \to H$ be a morphism of groups where $f$ is surjective and $|G|=pq$ where $p$ and $q$ are primes, then $H$ is abelian or $H \cong G$. Regard this is a well studied group I have never deal with a group like $G$. As $f$ is abelian we already got that $f(G)=H$ so in which case I can take lets say $h_{1}, h_{2} \in H$ arbitrary and I want to prove that $h_{1} * h_{2}=h_{2}* h_{1}$. As we have surjection of $f$ there are $g_{1}, g_{2} \in G$ such $f(g_{1})=h_{1}$ and $f(g_{2})=h_{2}$, also there is some $g \in G$ such $f(g)=h_{1}* h_{2}$ but still I cant do a lot from here as I dont know to use the hypothesis of $|G|=pq$.
Also any idea to prove in which other case we have that $f$ is an injection, I cant prove that $Ker(f)=\lbrace 0 \rbrace$.
This doesn't require touching elements at all. $G$ need not be abelian, but by the first isomorphism theorem we have that $H\cong G/K$ where $K$ is the kernel of $f$. In particular the order of $H$ divides the order of $G$. If $f$ is not an isomorphism, then $H$ has strictly smaller order. Since $|G|=pq$ it follows that $|H|\in \{p,q,1\}$, as these are the only positive integers smaller than $pq$ that divide $pq$. Groups of prime order and the trivial group are abelian.