# Thread: Group theory

1. ## Group theory

Let p and q be distinct primes. Suppose that H is a proper subset of the integers and H is a group under addition that contains 3 elements of the set { p, P+q, pq, p^q, q^p}
Determine the three elements in H.
Is it a. pq,p^q,q^p
or b. p+q,pq,q^p
c. p, pq, p^q.

I know that c is the right answer but I have not clue why. If H is a group then I would imagine that when I add p +pq I should get another element in H but i dont think so. I cant find an identity or inverse. Thanks

2. ## Re: Group theory

Originally Posted by jcir2826
Let p and q be distinct primes. Suppose that H is a proper subset of the integers and H is a group under addition that contains 3 elements of the set { p, P+q, pq, p^q, q^p}
Determine the three elements in H.
Is it a. pq,p^q,q^p
or b. p+q,pq,q^p
c. p, pq, p^q.

I know that c is the right answer but I have not clue why. If H is a group then I would imagine that when I add p +pq I should get another element in H but i dont think so. I cant find an identity or inverse. Thanks
Well we know that if $(a,b)=1$ then $a$ and $b$ will generate $\mathbb{Z}$. Since $H$ is a PROPER subset of $\mathbb{Z}$, then it cannot have such pairs of elements. Notice that $(p^q,q^p)=(p+q,q^p)=1$. So none of these pairs can be in $H$. Meanwhile notice that $pq$ and $p^q$ are generated by $p$. So $\langle p,pq,p^q\rangle=\langle p\rangle=\{n\cdot p:n\in\mathbb{Z}\}\subsetneq\mathbb{Z}$.

3. ## Re: Group theory

Originally Posted by jcir2826
Let p and q be distinct primes. Suppose that H is a proper subset of the integers and H is a group under addition that contains 3 elements of the set { p, P+q, pq, p^q, q^p}
Determine the three elements in H.
Is it a. pq,p^q,q^p
or b. p+q,pq,q^p
c. p, pq, p^q.

I know that c is the right answer but I have not clue why. If H is a group then I would imagine that when I add p +pq I should get another element in H but i dont think so. I cant find an identity or inverse. Thanks
A simple exercise: All subgroups of $\inline \mathbb{Z}$ are of the form $\inline n\mathbb{Z}= \{nx | x\in \mathbb{Z}\}$ for some natural number $\inline n$.
Only option (c) satisfies the above property.

4. ## Re: Group theory

in other words, here is what you have to do to check if your 3 answers could be all in some proper subgroup of Z:

a) pq could only be an element of pZ or qZ, since p and q are the only prime divisiors. p^q is in pZ, but not in qZ, so our 3rd element has to be in pZ. but q^p is not in pZ.

b) again, it's easier to start with pq. since q^p is in qZ, but not pZ, p+q would have to be in qZ. but p+q = qk implies p = q(k-1) is divisible by q, which is impossible.

c) p can only be in pZ (any strictly larger subgroup of Z must contain a divisor of p, which means it contains 1, which generates Z). as pq and p^q are both in pZ, we are done.

Z has an interesting property: subgroups in terms of addition, have properties related to multiplication. that is, mZ is a subgroup of nZ if and only if n divides m.

that is, studying ONLY the additive structure of Z, gives us information about the multiplicative structure of Z. this is why group theory is actually useful in number theory,

the lattice of additive subgroups of Z is the same as the lattice of divisiblilty (upside-down).

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# let p,q be distinct primes. then

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