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I was just doing some practice problems in my abstract algebra book trying to get a warm up this morning, but I found a GRE problem in the problem set and I don't know how to solve it. I've tried to think of examples of proper subgroups of integers under addition so I might get an intuition, but I was able to.

Let $p$ and $q$ be distinct primes. Suppose that $H$ is a proper subset of the integers that 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 three elements are in $H$.

The answer is $p,pq,p^{q}$. However, I am completely confused as how to get here. I thought I might know something, but I don't really have any reasons for why they might be, so I figured I don't actually know why.

Could someone show me the proof or reasoning why this is so?

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  • $\begingroup$ GRE exam means? $\endgroup$
    – Avinash N
    Jul 2, 2019 at 17:49

3 Answers 3

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The additive subgroups of ${\mathbb Z}$ are all of the form $\{kn: k \in {\mathbb Z}\}$ for an integer $n$. In other words $H$ consists of all multiples of some integer $n$. Here $n$ can't be $1$ or $-1$ since it is a proper subgroup.

The only subset of the $5$-member set that consists of multiples of a single integer $n \neq 1$ or $-1$ are $\{p,pq,p^q\}$, which are all multiples of $p$. Hence that's your answer.

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    $\begingroup$ Hey Zarrax, I appreciate your answer, but to be honest it is a little over my head. This question came up in the first section on groups. Is there a way you could go in a little more depth, or give a little less elegant proof with more of an explanation if you find time? Thank you very much $\endgroup$
    – Valentino
    Apr 21, 2015 at 13:53
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Let $K$ be a subgroup of $\mathbb{Z}$. If there are elements $a,b\in K$ such that $\gcd(a,b)=1$, then by Bezout's identity (Wikipedia link) we know $1\in K$, hence $n\in K$ for every $n\in \mathbb{Z}$, i.e. $K=\mathbb{Z}$.

Because $H$ is required to be a proper subgroup, there cannot be any two such elements in $H$.

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( $\Bbb{Z}$,+) is a cyclic group of infinite order. And it's known that every subgroup of $\Bbb{Z}$ will be also a cyclic group. Now as o( $\Bbb{Z}$ ) is infinite, it has only two generators, 1 and -1. As $\Bbb{H}$ is a subgroup of $\Bbb{Z}$ , it's cyclic. But it's generators can't be 1 or -1 , as $\Bbb {Z-H }$ ${\cap}$ $\Bbb{H}$ = ${\phi} $ i.e $\Bbb{H}$ ${\subset}$ $\Bbb {Z}$ . If p be the generator of $\Bbb{H}$ , $\Bbb{H}$ can be written in the form of , $\Bbb{H}$={pn: $n\in{\Bbb Z}\}$. Also if we call the given set 'S'; S={ ${p,pq ,p^q, q^p, p+q}$ }.

In the list of elements in $\Bbb{H}$ of S , p+q can be erased as $\Bbb{H}$ is a additive cyclic group. Then the remaining possibilities are,

  1. p, pq, ${p^{q}}$

  2. p, ${p^{q}}$ , ${q^{p}}$.

  3. p, pq, ${q^{p}}$

    As it is given that, gcd (p,q)= 1 and p,q are distinct primes , which suggests that p will not devide ${p^{q}}$ and ${q^{p}}$simultaneously and moreover p will not devide ${q^{p}}$ any way , i.e we won't be able to find out some $ n'\in{\Bbb Z}\}$ such that n'p= ${q^{p}}$ . And by this chain of argument we can erase (2) and (3) from the list . So $\Bbb{H}$ will contain exactly 3 elements of S , which are

${p,pq,p^q}$

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  • $\begingroup$ Please use MathJax to format your posts. $\endgroup$
    – jgon
    Mar 27, 2019 at 13:47

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