Relation between reduced rational numbers 
Suppose that we are given reduced rational numbers $\,\dfrac{a}{k},\
 \dfrac{b}{\ell},\ \dfrac{c}{q},\,$ i.e.
$\gcd(a,k)=\gcd(b,\ell)=\gcd(c,q)=1$ such that
$$\frac{c}{q}=\frac{a}{k}-\frac{b}{\ell}.$$
Then we have $\ q=k'\ell'e = {\dfrac{gk'l'}{f}}$ and $\,c=\dfrac{a\ell'-bk'}{f}$, where
$$g=\gcd(k,l),\,\ k'\,=\frac{k}g,\,\ \ell'=\frac{\ell}g,\,\ e=\frac{g}f,\ f=\gcd(a\ell'\!-bk',g).$$

I have some troubles to prove that $\,q=\ell'k'e.\,$ I was trying something like that: $$\frac{c}{q}=\frac{a\ell-bk}{k\ell}.$$ If we assume that $t=\gcd(a\ell-bk,k\ell)$, then $c=\dfrac{(a\ell-bk)}{t}$  and $q=\dfrac{kl}{t}$. Then I performed some manipulations but I did not reach the needed equality.
Can anyone show it please? Thank you!
 A: For simpler algebra, let $g = \gcd(k,\ell)$, which means $\ell = g\ell'$ and $k = gk'$. This gives that
$$\begin{equation}\begin{aligned}
\frac{a\ell - bk}{k\ell} & = \frac{g(a\ell' - bk')}{(gk')(g\ell')} \\
& = \frac{a\ell' - bk'}{gk'\ell'}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Next, using that $g = ef$, and $f = \gcd(a\ell' - bk',g) \; \to \; f \mid a\ell' - bk'$ meaning there's an integer $h$ such that $h = \frac{a\ell' - bk'}{f}$, we then get
$$\begin{equation}\begin{aligned}
\frac{a\ell' - bk'}{gk'\ell'} & = \frac{a\ell' - bk'}{efk'\ell'} \\
& = \frac{\left(\frac{a\ell' - bk'}{f}\right)}{ek'\ell'} \\
& = \frac{h}{ek'\ell'}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note from $f = \gcd(a\ell' - bk',g)$ with $a\ell' - bk' = hf$ and $g = ef$, we have that $\gcd(h,e) = 1$. Also, since $\gcd(a,k) = 1$, then $\gcd(a,k') = 1$, plus $\gcd(k',\ell') = 1$ (due to their definitions involving dividing by $\gcd(k,\ell)$), thus $\gcd(k',a\ell' - bk') = 1$, so $\gcd(k',h) = 1$. Similarly, $\gcd(\ell',h) = 1$. This means $\gcd(h, ek'\ell') = 1$, so \eqref{eq2A} is in the reduced rational form of $\frac{c}{q}$ with $c = h = \frac{a\ell' - bk'}{f}$ and $q = \ell'k'e$.
A: $\ \,\dfrac{c}q \,=\, {\dfrac{a}k-\dfrac{b}\ell} \,=\!\!\!\!\!\!\!\!\overbrace{\dfrac{a\ell -bk}{k\ell}}^{\!\textstyle\small {\rm cancel}\,\ g \!=\! (k,l) \Rightarrow }\!\!\!\!\!\!\!\!= {\dfrac{\color{#0a0}{a\ell' -bk'}}{\color{#0a0}g\:\!\color{#c00}{k'\ell'}}},\ $  for $\,\ \ell' = \dfrac{\ell}g,\,\ k' = \dfrac{k}g,\,\ (k',\ell')=1$
$\ \color{#c00}{k'}\,$ & $\,\color{#c00}{\ell'}$ are already coprime to $\,\color{#0a0}{a\ell'\! -bk'}$ (proof below), so to reduce  $\rm\color{#0a0}{RH}\color{#c00}{S}$ fraction it suffices by  $\small \rm\color{#0af}{EL}=$ Euclid's Lemma to cancel  $\:\!(\color{#0a0}{a\ell'\!-\!bk',g}) =:\! \color{darkorange}f,\,$ yielding $\,\bbox[6px,border:1px solid #c00]{\begin{align}c &= (a\ell'\!-bk')/ \color{darkorange}f\\ q &= gk'\ell'/ \color{darkorange}f,\,\text{ as claimed}\end{align}}$
Proof: $\,\ (\color{#c00}{k'},\:\!a\ell'\! -b\color{#c00}{k'})=\!\!\! \underbrace{(k',a\ell')\overset{\rm\color{#0af}{EL}}= 1}_{\small\textstyle  (k',a)\!=\!1\!=\!(k',\ell')}\!\!\!.\,$ By $\,k',\ell'$ symmetry $\,(\color{#c00}{\ell'},a\color{#c00}{\ell'}\! -bk')\!=\!1\,$ too.
