# If p and q are prime which elements are in the subgroup? (GRE question)

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?

• GRE exam means? Jul 2, 2019 at 17:49

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.

• 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 Apr 21, 2015 at 13:53

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 $H$.

( $$\Bbb{Z}$$,+) 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

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