Finding elements of a set that is itself a group under addition. Let p and q be distinct primes. 
Suppose that H is a proper subset of the integers and H is a group under addition that contains exactly three elements of the set  $\big\{p,p+q,pq,p^{q},q^{p}\big\} .$
Determine which of the following are the three elements in H.
I'm looking for a hint to do this question. A vital clue and the reasoning behind would be really helpful!
 A: Check each element to see what it generates and what it can't generate. If you end up with too many elements then you're out of luck for that generator, and if you end up with too few then you need to add another generator from the set. Try this until you get something with exactly 3 elements.
A: First, observe that the set $$S=\{p,p+q,p q,p^q,q^p\}$$ has exactly 5 members because $p,q$ are distinct primes. Second,observe $1\not \in H,$ else $H=Z$ and then $H$ contains all 5 members of $S.$ Third, observe that $p+q$ is co-prime to each of the other 4 members of $S.$ Therefore $p+q\not \in H.$ Proof:  If $p+q\in H,$ let $x\in H$ with $x\in S$ and $x\ne p+q.$ Then $A(p+q)$ and $B x$ belong to $H$ for all integers $A,B$, and there exist integers $A,B$ with $A(p+q)+B x=1$ (because $p+q$ and $x$ are co-prime), implying $1\in H.$ Fourth,we have $p\in H.$ Proof.If $p\not \in H$ then $S\cap H=\{p q,p^q,q^p\}.$ But $p^q$ and $q^p$ are co-prime so there are integers $A,B$ with $1=A p^q+B q^p\in H.$ Fifth,since $p\in H,$ every multiple of $p$ is in $H,$ so $p q$ and $p^q$ are in $H.$ And $q^p\not \in H.$ We conclude that $$H\cap S=\{p,p q ,p^q\}.$$ We also conclude that $H=Z p=\{p a :a\in Z\}$ because  (i) $ p\in H\implies Z p\subset H,$ and (ii) if $n\in Z\backslash Z p$  there are integers $A,B$ with $A n+B p=1,$ so $n\not \in H$.
