Let $H=\langle (3^9, 3^2, 3^7) \rangle$. The factor group $G:=\frac{\mathbb{Z}_{3^{12}} \times \mathbb{Z}_{3^{11}}\times \mathbb{Z}_{3^{10}}}{\langle (3^9, 3^2, 3^7) \rangle}$ is isomorphic to $\mathbb{Z}_{3^{12}} \times \mathbb{Z}_{3^{2}}\times \mathbb{Z}_{3^{10}}$ which is generated by $\{(1,0,0), (0,1,0), (0,0,1)\}$. I want to find an isomorphism between these two groups. What I tried is the following:
The order of $\chi_1=(1,0,0)+H $ is $3^{12}$ and order of $\chi_3=(0,0,1)+H $ is $3^{10}$. But order of $(0,1,0)+H$ is not $3^2$.
If we find an element of order $3^2$, say $\chi_2$ such that, any element $(a,b,c)+H$, say of $G$ can be generated by $\chi_1, \chi_2, \chi_3$, then the mapping $f$ defined by $$f(\chi_1)=(1,0,0), f(\chi_2)=(0,1,0), f(\chi_3)=(0,0,1) $$ is an isomorphism.
However, I could not manage to find an appropriate $\chi_2$. Although I found element of order $3^2$, for example, $(0,3^9, 0)+H$ which is of order $3^2$, but this cannot be taken as $\chi_2$. because if not, then there should be integers $\lambda_1, \lambda_2, \lambda_3 $ such that $$(a-\lambda_1, b-\lambda_2 3^9, c-\lambda_3 )=m(3^9, 3^2, 3^7)$$ for some integer $m$. But this leads to the following system of linear congruences \begin{align} a&\equiv \lambda_1[3^9]\\ b&\equiv \lambda_2 3^9[3^2]\\ c&\equiv \lambda_3[3^7] \end{align} which might not be true always.
What should I do? Any help will be appreciated.
Update1 I have made a mistake. The element $(0,3^9,0)+ H$ is actually $H$ itself, hence order is not $3^2$ but 1. However, I am still stuck in my old situation, how to find the desired generating element of order $3^2$.