Trying to show a set is a subgroup of $\mathbb{Q}$ I'm trying to show that for a fixed $n\in\mathbb{Z}^+$ the set of rational numbers whose denominators divide $n$ (call the set $H$) is a subgroup of $\mathbb{Q}$. If I let $a/b,c/d\in H$, then adding them gives $(ad+bc)/bd$. However, does $bd$ necessarily divide $n$? If I let $n=bk$ and $n=dl$ for appropriate $k,l\in\mathbb{Z}$, then I get $n^2=bdkl$. Not sure why this shows $bd$ divides $n$.
Don't use the formula
$$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$$
Instead, using the hypothesis, just express each fraction with a denominator of exactly $n$.

Explicitly, suppose $x \in H$. By definition of $H$, we can write $x = {\large{\frac{a}{b}}}$, where $a,b \in \mathbb{Z}$, and $b|n$. 

Then $b|n$ implies $n=bc$, for some $c \in \mathbb{Z}$, hence
$$x = \frac{a}{b} = \frac{ac}{bc} = \frac{ac}{n}$$
Thus, each element of $H$ can be expressed with a denominator of $n$.

It follows that $H$ is the set of rational numbers which can be expressed with a denominator of $n$.

From there, verifying the group axioms is easy.
Consider the homomorphism $\phi: \mathbb Q \to \mathbb Q$ given by $q \mapsto nq$.

Then $H = \phi^{-1}(\mathbb Z)$ and so is a subgroup.