A subgroups $K<G$ is *almost-malnormal* if $\forall g \in G \setminus K$, $K \cap gKg^{-1}$ is finite.

**Question:** Are there infinite groups in which all proper subgroups are almost malnormal?

Any such group $G$ must be fairly closed to being simple (since any normal subgroup is finite). Groups were all normal subgroups are finite (like Higmann's group) have a quotient which is simple (quotient by the maximal proper normal subgroup).

My initial question was: given an infinite simple group $G$ and an infinite subgroup $K$, is $K$ necessarily almost-malnormal?

A first try was to consider $Alt_{fin}$ (the group of finitely supported even permutations). But if $H$ is any infinite strict subgroup, then for any $\sigma \notin H$, there are infinitely many permutations in $H$ with support disjoint to $\sigma$. In other words there is an infinite subgroup $K <H$ with $\sigma K \sigma^{-1} = K$. So actually any infinite subgroup is quite far from being almost-malnormal.

I tried this out with next easy candidate: Higmann 4-generator 4-relator group $G = \langle a,b,c,d | bab^{-1} = a^2 , cbc^{-1} = b^2, dcd^{-1} = c^2, ada^{-1}= d^2 \rangle$. If $H$ is generated by three letters (say $a,b,c$) then $H \cap dHd^{-1} \supset \lbrace c^{2n} \rbrace_{n \in \mathbb{Z}}$.

*Remarks:* (added in Edit)

- $G$ is considered to be almost-malnormal in $G$ (whence "proper"). 
- Any finite subgroup is almost-malnormal. 
- When the group is torsion free, $K \cap gKg^{-1}$ is finite implies it is trivial (so the subgroup is malnormal).