9367 55

Forum for the GRE subject test in mathematics.
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yoyobarn
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Joined: Sun Dec 19, 2010 7:01 am

9367 55

Post by yoyobarn » Mon Dec 19, 2011 12:24 pm

Can someone elaborate more on this question?
(there is a brief explanation on http://www.mathematicsgre.com/viewtopic ... 7+55#p3839)

Let p and q be distinct primes. There is a proper subgroup J of the additive groups of integers which contains exactly three elements of the set {p,p+q, pq,p^q, q^p}. Which three elements are in J?

The answer is E.

However, from my basic knowledge of subgroups, the properties of closure, associativity (follows from group), identity and inverse must be fulfilled.

However, if we assume p is the identity, then how do the closure and inverse properties hold?

Thanks for clearing my misunderstanding.

Deltadesu
Posts: 4
Joined: Sun Dec 18, 2011 3:52 pm

Re: 9367 55

Post by Deltadesu » Tue Dec 20, 2011 5:14 am

The identity is 0, not p, because this is a subgroup of the additive group of integers, and the identity of the additive group of integers is 0. Hopefully that clears up your question.

In particular, the only nontrivial proper subgroups of the additive group of integers (Z,+) are of the form (nZ,+), where n is a positive integer. Eg, (2Z, +) is the group (..., -2, 0, 2, 4, ...) under addition. Clearly, (pZ, +) contains exactly p, pq, and p^q, so you're done.

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

Re: 9367 55

Post by yoyobarn » Tue Dec 20, 2011 5:37 am

Deltadesu wrote:The identity is 0, not p, because this is a subgroup of the additive group of integers, and the identity of the additive group of integers is 0. Hopefully that clears up your question.

In particular, the only nontrivial proper subgroups of the additive group of integers (Z,+) are of the form (nZ,+), where n is a positive integer. Eg, (2Z, +) is the group (..., -2, 0, 2, 4, ...) under addition. Clearly, (pZ, +) contains exactly p, pq, and p^q, so you're done.
Thanks for your clear explanation.

Ok, I think I understand. J contains other elements, other than the 3 elements.

In particular, J={...,-3p,-2p,-p,0,p,2p,3p,...}

I misunderstood the question and thought J had to have 3 elements, i.e. I thought |J|=3.



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