I want to know whether my approach to solving this sum is correct.
Suppose that $G$ is a finite abelian group of order $3^{4}\times 3^{10}$ and I want to show $G$ is isomorphic to $\mathbb{Z}_{81}\times \mathbb{Z}_{3^{10}}$. Suppose I show $G$ has at least one element of order $3^4$ and one element of order $3^{10}$, say $a, b$ respectively, such that $a\not\in \langle b\rangle$ and $b\not\in\langle a\rangle$, can I then state $G\simeq \mathbb{Z}_{81}\times \mathbb{Z}_{3^{10}}$? All I want to know is if this is the correct approach or not, and if not, what else I need to follow. Please help.
counterexample:
$G=\mathbb Z_{3^{13}}\times\mathbb Z_{3}$, now set $a=(3^3,0),b=(3^{9},1)$ and note that $o(a)=3^{10},o(b)=3^{4}$. Clearly we have that $b\not\in \langle a\rangle$ (has a one in the second component) and $a\not\in \langle b\rangle$ ($o(a)>o(b)$).