Solving Abstract Algebra Problem: 3 Elements of {p,p+q,pq, p^q,q^p}

[lhuyvn]

Hi again,

I'm preparing for GRE Math. I regconize that Abstract Algebra is the most tough. Here is one of its problem, Any suggestion!.

Thank you in advance.

Let p and q be distinct primes. There is a proper subgroup J of the additive group of integers which contains exactly three elements

of the set {p,p+q,pq, p^q,q^p}, which three elements are in J.

A. pq, p^q, q^p
B. p+q,pq, p^q
C. p,p+q, pq
D. p,p^q,q^p
E. p, pq,p^q

 

[AKG]

There is a proper subgroup J of the additive group of integers

So doesn't this mean that the group operation for J is regular addition? And since it is an additive group, it should satisfy:
1. If A, B are in J, then A+B in J
2. If A, B, C are in J, then (A+B)+C = A+(B+C)
3. There is an I such that A+I = I+A = A
4. There is an inverse, A' for each A such that A' + A = A + A' = I

I've probably interpreted something wrong (or you haven't given enough information) because this doesn't seem possible. Perhaps it's not regular additions we should be looking at?

 

[Icarus]

Every non-trivial additive subgroup of the integers is generated by its least positive element. So J = {ma | m in Z} for some a. Out of the 5 elements in the given set, you need to choose a set of 3 all of which are multiples of the same number, and such that the other two elements are not multiples of this number.

Note that p, pq, p^q are all multiple of p.
On the other hand, since p and q are relatively prime, p+q and q^p are not multiples of p.

Hence E is your answer.