Hw4 Problem 10

### Problem 10

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 are the three elements in \$H\$.

(a) \$pq,p^{q},q^{p}\$
(b) \$p+q,pq,p^{q}\$
(c) \$p,pq,p^{q}\$
(d) \$p,p+q,pq\$
(e) \$p,p^{q},q^{p}\$

Solution

Use relative primality: if any two elements are relatively prime, we may write a Z-linear combination of this pair as \$1\$. Then, we can easily generate the elements not picked in the list.

For us, we can do this with all choices but part (c). (All elements in part (c) are multiples of \$p\$; hence you can't generate \$q\$ because \$gcd(p,q) = 1\$.)

So, only (c) can form a group \$H\$.

As for the other parts:

(a) Since \$gcd(p,q) = 1\$, \$gcd(p^{q},q^{p}) = 1\$. Thus, \$1\$ is in \$H\$. Since \$1\$ is in \$H\$, any multiple of \$1\$, such as the elements not used from your given set are also in \$H\$.

(b) Since \$gcd(p,q) = 1\$, \$gcd(p+q, p) = 1\$ and thus \$gcd(p+q, p^{q}) = 1\$. Thus, \$1\$ is in \$H\$.

(d) Since \$gcd(p,q) = 1\$, \$gcd(p+q, p) = 1\$. Thus, \$1\$ is in \$H\$.

(e) See part (a).