# Another question.

1. Jun 8, 2004

### 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!!!.

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

2. Jun 8, 2004

### AKG

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?

3. Jun 8, 2004

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