Level 24, part 1

Level 24
$$\begin{align} j_{24A}(\tau) &= \left(\frac{d_4^2\, d_{12}^2}{d_2\, d_6\, d_8\, d_{24}}\right)^3=\sqrt{j_{12A}(2\tau)}\\ &=j_{24D}(\tau)+\frac{4}{j_{24D}(\tau)}\\ &=j_{24H}(\tau)+\frac{1}{j_{24H}(\tau)}\\ \color{blue}{j_{24B}(\tau)} &= \left(\frac{d_2\, d_4\, d_6\, d_{12}}{d_1\, d_3\, d_8\, d_{24}}\right)^2\\ \color{red}{j_{24C}(\tau)} &= \left(\frac{d_1^2\, d_6\, d_8^2\, d_{12}}{d_2\, d_3^2\, d_4\, d_{24}^2}\right)\\ j_{24D}(\tau) &= \left(\frac{d_2\, d_6}{d_8\, d_{24}}\right)=\sqrt{j_{12E}(2\tau)}\\ j_{24E}(\tau) &= \left(\frac{d_{12}^2}{d_6\, d_{24}}\right)^4=\sqrt{j_{12D}(2\tau)}\\ j_{24F}(\tau) &= \left(\frac{d_8\, d_{12}}{d_4\, d_{24}}\right)^3 = \sqrt{j_{12F}(2\tau)}\\ j_{24G}(\tau) &= \left(\frac{d_4\, d_8}{d_{12}\, d_{24}}\right) = \sqrt{j_{12G}(2\tau)}\\ j_{24H}(\tau) &= \left(\frac{d_6\, d_8}{d_2\, d_{24}}\right)^2=\sqrt{j_{12H}(2\tau)}\\ \color{red}{j_{24I}(\tau)} &= \left(\frac{d_2\, d_3^2\, d_8^2\, d_{12}}{d_1^2\, d_4\, d_6\, d_{24}^2}\right)\\ j_{24J}(\tau) &= \;\left(\frac{d_{12}}{d_{24}}\right)^2\;=\;\sqrt{j_{12J}(2\tau)}\\ \end{align}$$ The "important" functions (that aren't just square roots) are $j_{24C}$ and $j_{24I}$. First define the auxiliary non-monster functions, $$\begin{align} U &=\left(\frac{d_1\,d_6\,d_8\,d_{12}}{d_2\,d_3\,d_4\,d_{24}}\right)^4\\ V &=\left(\frac{d_2^2\,d_3\,d_8\,d_{12}^2}{d_1\,d_4^2\,d_6^2\,d_{24}}\right)^4\quad\quad \end{align}$$ I. Let $a_1=\color{red}{j_{24C}},\;m = 1,\;n=3,$ then,
$$\begin{align} \color{blue}{j_{24B}} &= a_1+\frac{mn}{a_1}+(m+n)\\ j_{12G} &= a_1+m+\frac{m(m-n)}{a_1+m}-(2m-n)\\ U &= a_1+n+\frac{n(m-n)}{a_1+n}+(m-2n) \end{align}$$ as well as, $$j_{12C} =  j_{24B}+\frac{(m-n)^2}{j_{24B}}-2 =  j_{12G}+\frac{n^2}{j_{12G}} =  U+\frac{m^2}{U}+4$$
II. Let $a_2=\color{red}{j_{24I}},\;m = 1,\;n=-1,$ then,
$$\begin{align} \color{blue}{j_{24B}} &= a_2+\frac{mn}{a_2}+(m+n)\\ j_{12F} &= a_2+m+\frac{m(m-n)}{a_2+m}-(2m-n)\\ V &= a_2+n+\frac{n(m-n)}{a_2+n}+(m-2n) \end{align}$$ as well as, $$j_{12C} =  j_{24B}+\frac{(m-n)^2}{j_{24B}}-2 =  j_{12F}+\frac{n^2}{j_{12F}} =  V+\frac{m^2}{V}-4$$ These two families imply the two linear relations, $$j_{12C}+2\color{red}{j_{24C}} = j_{24B}+j_{12G}+U-2\\j_{12C}+2\color{red}{j_{24I}} = j_{24B}+j_{12F}+V-2$$  
However, there is one linear relation using only monster functions,
$$j_{12C}-j_{12F}-j_{12G}+j_{12E}-2j_{12I} = 2\big(\color{blue}{j_{24B}}-\color{red}{j_{24C}}-\color{red}{j_{24I}}-3\big)$$ Note that,
$$j_{12C}=j_{12F}+\frac{1}{j_{12F}} = j_{12G}+\frac{9}{j_{12G}}\\ j_{12E} = j_{12I}-\frac{3}{j_{12I}}-2\quad$$