This is question is related to factor group. I am trying my best to provide details as much as possible. All I need to know is whether I have solved the sums correctly or not.
Let $\chi=(3,2)+\langle (5^6, 5^4)\rangle\in G:=\frac{\mathbb{Z}_{5^8}\times \mathbb{Z}_{5^6}}{\langle (5^6, 5^4)\rangle}$. Here the order of the group $G$ is $5^{12}$. Hence by lagrange's theorem order of $\chi$ is $5^r$ where $1\leqslant r\leqslant 12$, since $\chi$ is an non-identity element. Since $\langle (5^6, 5^4)\rangle$ is the identity element of $G$, there must exist an integer $m$ such that $5^r (3,2)=m(5^6, 5^4)$ and that shows \begin{align} 3.5^r\equiv 5^6 m[5^8]\\ 2.5^r\equiv 5^4 m[5^6] \end{align} The first equation reduces to $3.5^{r-6}\equiv m[5^2]$ and the second one becomes $2.5^{r-4}\equiv m[5^2]$. Hence $3.5^{r-6}\equiv 2.5^{r-4}[5^2]$ i.e. $3.5^r\equiv 2.5^{r+2}[5^8]$. The smallest positive $r$ that will satisfy this last equation is $r=8$ and so $|\chi|=5^8$. Hence $5^8(3,2)=(0,0)=5^2(5^6, 5^4)$ gives $m=5^2$.
Please let me know whether I have done it correctly. Any help is appreciated. Thanks in advance.