1. ## question

P and q are two distinct primes. H ia s proper subset of integers and also it is a group under addition. another set A={p,p+q,pq,p^q,q^p}. now H has three elements and those three elements are from A. which are those three elements? please reply in detail

2. Originally Posted by luckylawrance
P and q are two distinct primes. H ia s proper subset of integers and also it is a group under addition. another set A={p,p+q,pq,p^q,q^p}. now H has three elements and those three elements are from A. which are those three elements? please reply in detail
I presume you mean $H \cap A$ contains 3 elements, not $H$ contains three elements (as subgroups of $\mathbb{Z}$ are either trivial or contain infinitely many elements).

Now, subgroups of $\mathbb{Z}$ are all of the form $n\mathbb{Z} = \{ni; i \in \mathbb{Z}\}$ (why?). Thus, there exists $n \in \mathbb{Z}$ such that $H=$ ( $H$ is the subgroup generated by $n$).

Basically, this all translates as "the elements you are looking for are elements $a, b, c \in A$ such that $gcd(a, b, c) \neq 1$" (not equal to 1 as H is a proper subgroup).

Can you find three such elements?

3. first of all thank you for replying. but question in the book goes like this only..that's why i am unable to get the solution

4. Originally Posted by luckylawrance
first of all thank you for replying. but question in the book goes like this only..that's why i am unable to get the solution
Well, what bit don't you understand?

5. the solution given in book is {p,p+q,p^q}. but how is it possible that sum of any two of these three elements also lies in the group?

6. Originally Posted by luckylawrance
the solution given in book is {p,p+q,p^q}. but how is it possible that sum of any two of these three elements also lies in the group?
I believe it should be pq, not p+q (if my interpretation of the question, which I posted earlier, is correct).