Which Elements Belong in H? [SOLVED] Question about a Subset of Z

[DavidWhitbeck]

[SOLVED] Question about a subset of Z

So I was working on exercises out of Gallain's Algebra book and it looks like it doesn't actually have an answer! So of course I disagree with the answer in the back of the book, maybe I'm missing something.

Context Only provided theory is the definition of a group (chapter 2 in Gallain).

Problem

(From the GRE Practice Exam) 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 $$\{ p,p+q,pq,p^q,q^p \}$$. Determine which of the following are the three elements in H.
(a) $$pq, p^q, q^p$$
(b) $$p+q,pq,p^q$$
(c) $$p,p+q,pq$$
(d) $$p,p^q,q^p$$
(e) $$p,pq,p^q$$

My work
The identity for the addition operator is 0. If H is a group it must contain the identify as one of the elements. That implies that one of those four elements is 0.
(a) It's not p because p is prime, and 0 is not prime by definition.
(b) Similarly, it's not q.
(c) If $$p^q = 0 \Rightarrow p = 0$$ but p is not 0, so it's not $$p^q$$.
(d) Similarly, it's not $$q^p$$.
(e) If $$p + q = 0$$ then either p or q is negative. Negative integers are not prime by definition, so it's not p+q.

I have just shown that none of the elements are the identity, and so H is not a group and the problem is ill-posed.

Correct Solution (back of the book) (e)

Where have I gone wrong? Before you ask, I didn't type it wrong, I reproduced it exactly as it appeared in the book.

 

[tiny-tim]

Hi David!

H is a proper subset of the integers and H is a group under addition that contains exactly three elements of the set { , , , , }

I would read that as meaning that the group addition is the ordinary addition of integers, and that H contains infinitely many elements, but only three from the specified set.

 

[DavidWhitbeck]

Oh thanks Tim! You're right, and my misreading was enough to completely throw me off. Now I see it, I don't need 0 to be in there and no longer fixated around that I see that pq and p^q can all be built from adding p successive times and so it's (e). And I was making the problem harder than it was supposed to be.

 

[mathwonk]

the language "contains" is misleading, but for your interpretation to have been correct, it probably would have said "consists of".