$\mathcal Z$-transform, ROC of a system from dividing to others TRUE / FALSE

Given three systems with rational $\mathcal Z$-transform. Systems A and B are not stable with $\mathcal Z$-transform $H(z)$, $G(z)$ respectively. A and B have no common poles. System's C $\mathcal Z$-transform is $F(z)=H(z)/G(z)$, then C is also unstable?

False:

\begin{align}
H(z)&amp;= \frac{1}{1-2z^{-1}}: \textrm{ROC} \quad\lvert z\rvert&gt;2\\
G(z)&amp;= \frac{1}{1-3z^{-1}}: \textrm{ROC} \quad \lvert z\rvert&gt;3\\
F(z)&amp;= \frac{1-3z^{-1}}{1-2z^{-1}}: \textrm{ROC} \quad \lvert z\rvert&lt;2
\end{align}

It is the answer, but doesn't $F(z)$ ROC must be like that of $H(z$), or the division allows to choose the ROC I want?
I agree with your doubts. The $\mathcal{Z}$-transform

$$\hat{G}(z)=1/G(z)=1-3z^{-1}$$

has the ROC $|z|&gt;0$. The ROC of the multiplication of $H(z)$ and $\hat{G}(z)$ must be the intersection of the ROCs of $H(z)$ and $\hat{G}(z)$ (unless there occur pole-zero cancellations, which is not the case here). Since the ROC of $H(z)$ is $|z|&gt;2$, the ROC of $F(z)=H(z)\hat{G}(z)$ must also be $|z|&gt;2$.

Another way to think about it is that since $H(z)$ and $\hat{G}(z)$ are both causal, also $F(z)$ must be causal, i.e., its ROC must be outside a circle in the complex plane.