Abstract Algebra group problem.

  • #1
WHOAguitarninja
20
0
ProbelmLet p and q be distinct primes. Suppose that H is a porper 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



Homework Equations


The answer is e, as it's given in the book.



The Attempt at a Solution


My basic strategy here was to try and show it's only closed under addition for e. It's easy to see that adding p to itself q times yields pq, however the same cannot be said for p^q. This didn't seem satisfactory then. My next idea, the one that MIGHT be right, but I'm very much not sure, is that e contains the only terms that are integer factors of p. You know p, pq, and p^q are integers when divided by p, and furthermore they are the only ones that are (since p and q are prime). However, I don't know if this is even signficant, and I certainly don't see the signifigance under a group who's operation is addition.

EDIT - Actually I suppose it's significant in that being an integer multiple it can be seen as "P+P+P+..." an integer number of times. Is that all there is to it? It seems a bit of an odd problem if so.
 
Last edited:

Answers and Replies

  • #2
AKG
Science Advisor
Homework Helper
2,566
4
Hint 1: H has to be a proper subset of Z
Hint 2: If x and y are coprime, then there exists integers a and b such that ax + by = gcd(x,y)
 
  • #3
WHOAguitarninja
20
0
I've been fiddling for a little while and can't seem to come up with anything satisfactory. The crux of the problem seems to be that the items in the group must be integers, however I can't seem to find the relevance of your second hint. Having fiddled with it, it doesn't seem to get me anywhere.

Is it not important that only the terms in e can have p factored out and have the resulting term still be integer valued?
 
  • #4
AKG
Science Advisor
Homework Helper
2,566
4
I've been fiddling for a little while and can't seem to come up with anything satisfactory. The crux of the problem seems to be that the items in the group must be integers, however I can't seem to find the relevance of your second hint. Having fiddled with it, it doesn't seem to get me anywhere.

Is it not important that only the terms in e can have p factored out and have the resulting term still be integer valued?
Yes, that's good. Since p, pq, and pq are integer multiples of p, they can appear in the subgroup of Z consisting of all integer multiples of p, which is indeed a proper subgroup. Now with all the other options, a-d, you can't factor out p from each of the three elements. So p is never a common factor of the elements in a-d. But you need to show that for each of a-d, the three elements have no common factors at all.

The point of my hint was that if in each of a-d, you can find a pair of elements that are coprime, e.g. in b, pq and p+q are coprime. Therefore, there exist integers a and b such that a(pq) + b(p+q) = 1. But then 1 would be in H, and hence H would be all of Z, contradicting Hint 1.
 
  • #5
WHOAguitarninja
20
0
Yes, that's good. Since p, pq, and pq are integer multiples of p, they can appear in the subgroup of Z consisting of all integer multiples of p, which is indeed a proper subgroup. Now with all the other options, a-d, you can't factor out p from each of the three elements. So p is never a common factor of the elements in a-d. But you need to show that for each of a-d, the three elements have no common factors at all.

The point of my hint was that if in each of a-d, you can find a pair of elements that are coprime, e.g. in b, pq and p+q are coprime. Therefore, there exist integers a and b such that a(pq) + b(p+q) = 1. But then 1 would be in H, and hence H would be all of Z, contradicting Hint 1.


Thanks. That makes perfect sense. I wasn't thinking more along the lines of showing a-d were not groups than were not subgroups. I suppose it then makes sense that I could never explain that they weren't, because they are! Thanks alot, that really helped.
 

Suggested for: Abstract Algebra group problem.

Replies
5
Views
116
Replies
6
Views
553
  • Last Post
Replies
1
Views
588
Replies
1
Views
888
  • Last Post
Replies
6
Views
860
  • Last Post
Replies
4
Views
428
Replies
3
Views
582
  • Last Post
2
Replies
38
Views
4K
  • Last Post
Replies
10
Views
654
Top