Prove that both $R_p$ and $R^p$ are abelian groups under ordinary addition of rationals.

Question

Q: Let $p$ be a fixed prime. Let $R_p$ be the set of all those rational numbers whose denominator is relatively prime to $p$. Let $R^p$ be the set of rationals whose denominator is a power of $p\ (p^i, i\geq 0)$. Prove that both $R_p$ and $R^p$ are abelian groups under ordinary addition of rationals.

A:

In case of $R_p$:

1. $\forall a_1/b_1,a_2/b_2,a_3/b_3\in R_p=\{a/b: (b,p)=1\}; (a_1/b_1+a_2/b_2)+a_3/b_3=\dfrac{a_1b_2+a_2b_1}{b_1b_2}+\dfrac{a_3}{b_3}=\dfrac{a_1b_2b+a_2b_1b_3+a_3b_3b_2}{b_1b_2b_3}=\dfrac{a_1}{b_1}+\dfrac{a_2b_1b_3+a_3b_1b_2}{b_1b_2b_3}=\dfrac{a_1}{b_1}+(\dfrac{a_2}{b_2}+\dfrac{a_3}{b_3})$
   Semigroup ✓
2. $\forall a/b\in R_p, \exists 0/b\in R_p; a/b+0/b=0/b+a/b=a/b$
   Monoid ✓
3. $\forall a/b\in R_p, \exists (-a/b)\in R_p; a/b+(-a)/b=(-a)/b+a/b=0/b.$
   Group ✓
4. $\forall a_1/b_1,a_2/b_2\in R_p; a_1/b_1+a_2/b_2=\dfrac{a_1b_2+a_2b_1}{b_1b_2}=\dfrac{a_2b_1+a_1b_2}{b_1b_2}=a_2/b_2+a_1/b_1$
   Abelian Group ✓

In case of $R^p$:

1. $\forall a_1/p^i,a_2/p^j,a_3/p^k\in R^p=\{a/p^i : i\geq 0\}; (a_1/p^i+a_2/p^j)+a_3/p^k=\dfrac{a_1p^jp^k+a_2p^ip^k+a_3p^ip^j}{p^ip^jp^k}=a_1/p^i +(a_2/p^j +a_3/p^k)$
   Semigroup ✓
2. $\forall a/p_i\in R^p, \exists 0/p^1 \in R^p; a/p^i +0/p^1=0/p^1 +a/p^i=a/p^i$
   Monoid ✓
3. $\forall a/p^i \in R^p, \exists (-a)/p^i \in R; a/p^i+(-a)/p^i=(-a)/p^i+a/p^i=0/p^1.$
   Group ✓
4. $\forall a_1/p^i,a_2/p^j\in R^p; a_1/p^i+a_2/p^j=\dfrac{a_1p^j+a_2p^i}{p^ip^j}=\dfrac{a_2p^i+a_1p^j}{p^ip^j}=a_2/p^j+a_1/p^i.$
   Abelian Group ✓