Question about a subset of Z

  • #1
[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 [tex]p[/tex] and [tex]q[/tex] 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 [tex]\{ p,p+q,pq,p^q,q^p \}[/tex]. Determine which of the following are the three elements in H.
(a) [tex]pq, p^q, q^p[/tex]
(b) [tex]p+q,pq,p^q[/tex]
(c) [tex]p,p+q,pq[/tex]
(d) [tex]p,p^q,q^p[/tex]
(e) [tex]p,pq,p^q[/tex]
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 [tex]p^q = 0 \Rightarrow p = 0[/tex] but p is not 0, so it's not [tex]p^q[/tex].
(d) Similarly, it's not [tex]q^p[/tex].
(e) If [tex]p + q = 0[/tex] 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.
 

Answers and Replies

  • #2
tiny-tim
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Hi David! :smile:
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. :smile:
 
  • #3
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.
 
  • #4
mathwonk
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the language "contains" is misleading, but for your interpretation to have been correct, it probably would have said "consists of".
 

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