Does set of rational numbers with odd denominators form a group with binary operation of $+$?

Question

Does set of rational numbers with odd denominators form a group with binary operation of $+$?

I think no, because it doesn't have an identity element since $0$ is not in the set because it doesn't have a denominator.

Can I say $0$ does not have a denominator, so it doesn't have an odd denominator?

Answer 1

Yes, it forms a group. It has an identity element, which is $0\left(=\frac01\right)$.

Answer 2

Let $O\subset \mathbb Q$ denote the set of rational numbers such that the denominator is odd when written as a fraction where the numerator and denominator are relatively prime. Note that $\mathbb Z\subset O$, in particular $0\in O$ (note that even though $0=\tfrac{0}{n}$ is true for all integers $n>0$, the pair $(0,n)$ is relatively prime only when $n=\pm 1$, so the definition still works in this case).

You ask whether $O$ is a subgroup of $\mathbb Q$ under addition. And this is true, since $$ \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd} $$ has $bd$ odd if $b,d$ are both odd, so if $(a,b)=(c,d)=1$ then since $(ad+bc,bd)=1$ it follows that the sum of two elements of $O$ is also in $O$.

EDIT: It's been pointed out in the comments that the previous sentence is incorrect: it need not be the case that $(ad+bc,bd)=1$ under these hypotheses. However, that claim is actually irrelevant to the argument! After reducing the fraction, the denominator will (by definition) be a divisor of $bd$ which is odd. Since all divisors of an odd number are also odd, the claim follows.

Answer 3

The family $\Bbb Q_{(2)}:=\{\frac{a}{b}:a,b\in\Bbb Z\;\land\;2\nmid b\}$ of rationals "with odd denominator" is an integral domain, a subring of $\Bbb Q$ but not a field as $\frac{3}{4}\notin\Bbb Q_{(2)},\frac{4}{3}\in\Bbb Q_{(2)}$.

In particular, $\Bbb Q_{(2)}=\Bbb Z[\frac{1}{3},\frac{1}{5},\frac{1}{7},\frac{1}{11},\dots]$ and for each prime $p>2$ we have $\Bbb Q_{(p)}:=\{\frac{a}{b}:a,b\in\Bbb Z\;\land\;p\nmid b\}=\Bbb Z[\frac{1}{2},\dots]$ adjoining to $\Bbb Z$ all prime reciprocals except $\frac{1}{p}$. To see this simply note that $\frac{ab'-ba'}{bb'},\frac{aa'}{bb'}\in\Bbb Q_{(p)}$ is true whenever $\frac{a}{b},\frac{a'}{b'}\in\Bbb Q_{(p)}$ where $2\nmid b,b'$.