I'm trying to prove that $\langle 2 \rangle$ is not isomorphic to $\langle 2 ,3 \rangle$ as subgroups of $(\mathbb{Q}_{>0},\cdot)$.
I know that cyclic-ness is preserved under isomorphisms, so I imagine I want to show that $\langle 2,3 \rangle$ is not cyclic. Below is the start of a proof by contradiction, but I'm not sure where to take it from here.
Assuming $\langle 2,3 \rangle$ is cyclic means there exists $a \in \langle 2,3 \rangle$ such that $\langle 2,3 \rangle = \langle a \rangle$. Then there exists $k,j \in \mathbb{Z}$ such that $a^k = 2$ and $a^j = 3$. Then $a = \sqrt[k]{2} = \sqrt[j]{3}$, $a^{k+j} = 6$, and $a^{k-j} = {2 \over 3}$.
I can't figure out where to find a contradiction. I've seen and I understand this proof, but I'm having a hard time adapting what's done there to this setting.
Any hints would be greatly appreciated.
You are in the right track. Just show that the subgroup generated 2 and 3 cannot be generated by a single element. This canbe done by showing it is isomorphic to the subgroup of the plane consisting of all integer points.
The isomorphism sends the integer point $(m, n)$ to $2^m 3^n$. This is onto the subgroup, clearly. It is injective by fundamental theorem of arithmetic.
Of course one has to show that the group of integer points in the plane is not generated by a single element. This is easier than the original problem.