Determination of order of cosets in a factor group of a finite abelian group. Consider the group $G=\frac{\mathbb{Z}_{3^{10}}\times \mathbb{Z}_{3^7} }{\langle (3^2, 3^3)\rangle}$. Let $a=(1,0)+\langle (3^2, 3^3)\rangle$ and $b=(0,1)+\langle (3^2, 3^3)\rangle$. Since the order of the group $G$ is $3^9$, so order of $a$ and $b$ must be a divisor of $3^9$.
I am willing to find $|a|$ and $|b|$. Assume $\alpha=|a|$. Then
$\alpha(1,0)=m(3^2, 3^3)$ for some positive integer $m$ and so $\alpha\equiv 3^2 m[3^{10}], 0\equiv 3^3 m[3^7]$.
The second congruence yields $m=3^4 t$ where $t$ is chosen to be non-negative integers. Hence $\alpha\equiv 3^6 t[3^{10}]$. Do not understand what to do next to derive $\alpha$.
For the determination of order of $(0,1)+\langle (3^2, 3^3)\rangle$, let $\beta$ be its order. Then $\beta(0,1)=m'(3^2, 3^3)$ for some positive integer $m'$. So $0\equiv 3^2 m'[3^{10}], \beta\equiv 3^3 m'[3^7]$. The first case gives $m'=3^8 t, t\geqslant 1$ and so the second equation becomes $\beta\equiv 3^3(3^8 t')\equiv 0[3^7]$. How come this arrived ? I am really confused here.
Please help what to be done here.

Let $G_1=\Bbb{Z}_{3^{10}} $,$G_2=\Bbb{Z}_{3^{7}}$ , $H=\langle (3^2, 3^3)\rangle$
Then $|3^2|_{G_{1}}=3^8$ , $|3^3|_{G_{2}}=3^4$
Let $|a|=n$ where $a=(1,0)+H$
Then $n$ is the least positive integer such that $na=n(1, 0) +H=H$
$(n, 0) +H=H$ implies $(n, 0) \in H$
Hence $n$ is the least positive integer such that $(n, 0) \in H$
$|3^3|_{G_{2}}=3^4 \implies 3^7=0$.
Hence $3^4(3^2, 3^3) =(3^6, 0) \in H$
$3^6$ is the least positive integer in the first component such that $(3^6, 0) \in H$ implies $n=3^6$

 $|(1, 0) +H|=3^6$


Edit:
Let $|(0, 1) +H|=m$
Then $m$ is the least positive integer such that $(0, n) \in H$
$3^7$ is the least positive integer such that $(0, 3^7) =(0,0)\in H$.

$|(0, 1) +H|=3^7$

