Suppose $\mathcal T$ and $\mathcal S$ are two compatible Hausdorff topologies on a group $G$ and $\mathcal R$ is a maximum compatible topology on $G$ with $\mathcal R \subseteq \mathcal T\cap \mathcal S$.


Is $(G,\mathcal R)$ Hausdorff?



Hausdorffness inheritance in topological groups
It seems the answer is no. Basically, the $p$-adic and Archimedean topologies on $\mathbb{Q}$ are incompatible enough that the maximal compatible topology contained in both of them is indiscrete. Here is an outline of the proof (I'm heading to bed so I haven't filled in the details):


$\bullet$ Any subset of $\mathbb{Q}$ which is open for both topologies must be dense for both topologies. Therefore, all elements of $\mathcal{R}$ are dense. 


$\bullet$ If $\mathcal{R}$ is compatible with the group structure, the complement of the antidiagonal in $\mathbb{Q}\times \mathbb{Q}$ must be open. Since this complement contains e.g. $(1,1)$, it must contain a set of the form $U\times U$ for some $U\in \mathcal{R}$.


$\bullet$ This is impossible since $U\cap U^{-1}\neq \emptyset$ by density of $U$. 


