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Consider the following problem:

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 $J$?

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

Here are my questions:

  • What properties of ${\bf Z}$ does one need to use here?
  • How to solve the problem above?
  • Can one generalize this problem with changing the condition "exactly three"?
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1  
Concerning "generalizing", $J$ can't have exactly four; exactly five is silly; it can have exactly one, but there's more than one correct answer; looks like it's impossible to have exactly two. –  Gerry Myerson Jun 30 '11 at 4:17
    
Note that J may contain elements other than the "exactly three" elements of the given set...That is, I don't think the question is stating that J has exactly three elements, period. –  amWhy Jun 30 '11 at 4:42
    
@amWhy, I don't know if your comment was in response to mine. If it was, I was using "exactly four" as shorthand for "exactly four of those five elements, and possibly some other elements, too." If your comment was directed to OP then, in the words of Emily Litella (en.wikipedia.org/wiki/Emily_Litella), never mind. –  Gerry Myerson Jun 30 '11 at 5:33
    
Sorry, @Gerry: it was directed to OP. I understood you perfectly well! –  amWhy Jun 30 '11 at 5:36

2 Answers 2

up vote 8 down vote accepted

You need to use the fact that if $a,b$ are relatively prime then they generate all integers.

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That's a bit too complicated, and isn't really the easiest way of doing it. Finding the three elements with a common divisor is much quicker and easier... –  user1729 Jun 30 '11 at 8:51
    
Hmm, Bezout's identity. –  Jack Jul 6 '11 at 17:09

You don't need any properties of $\mathbb{Z}_n$. What you need is the fundamental theorem of arithmetic (unique prime factorization of integers) and the following property of the additive group $\mathbb{Z}$: the subgroups of $\mathbb{Z}$ are precisely the subsets $n\mathbb{Z}$ $(n\in\mathbb{Z})$. To prove this fact, show that if $a$ and $b$ are in your subgroup, then $\gcd(a,b)$ is in your subgroup.

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