Let $G:=\frac{\mathbb{Z}_{3^{15}}\times \mathbb{Z}_{3^{12}}}{\langle (3^7.2, 0)\rangle}$. The order of $G$ is $3^{19}$. Consider $\chi:=(3^4.7, 0)+K$ where $K=\langle (3^7.2, 0)\rangle$. To find the order of $\chi$.
I have attempted the solution in the following way. Please guide me if this is correct. Since $\chi\in G$, by Lagrange's theorem, order of $\chi$ is $3^r$ for some $r>0$. Now $|3^4.7|=3^{11}$ in $\mathbb{Z}_{3^{15}}$. Therefore $$3^{11}(3^4.7, 0)=(3^{15}.7, 0)=(0,0).$$ On the other hand, $3^8(3^7.2, 0)=(0,0)$. Thus $$3^{11}(3^4.7, 0)=(0,0)=3^8(3^7.2, 0)\in K$$ which shows $3^{11}((3^7.7, 0)+K)=K$ i.e. desired order of $\chi$ is $3^{11}$.
Is this correct? Please suggest. Thanks in advance.
Update1: I believe I have made mistake in my calculation. Since $$3^3(3^4.7, 0)=(3^7.7, 0)=2^{-1}.7(3^7.2, 0)\in K$$ therefore $3^3$ is the smallest positive integer for which $3^3(3^4.7,0)\in K$. Thus desired order is $3^3$. Is it correct?