Just to reiterate standard facts about Frobenius groups: a finite group $G$ is said to be a Frobenius group if $G$ contains a proper non-identity subgroup $H$ such that $H \cap gHg^{-1} = 1$ for all $g \in G \backslash H.$ This hypothesis is exactly equivalent to the permutation-group-theoretic definition of Frobenius groups: these are finite permutation groups $G$ in which no non-identity element fixes more than one point of a given set $\Omega$ on which $G$ acts transitively. Given such an action on a set $\Omega,$ just take $H$ to be a point stabilizer.
Conversely, given a subgroup $H$ with $H \cap gHg^{-1} = 1$ for all $g \in G \backslash H,$ then the usual action of $G$ by right translation on the set $\Omega$ of right cosets of $H$ in $G$ gives a permutation action in which no non-identity element fixes more that one point.
 The fundamental theorem of Frobenius is that given a Frobenius group $G$ with such a subgroup $H,$ there is a normal subgroup $K$ with $G = KH$ and $K \cap H = \{1 \}.$ This translates the problem of finding Frobenius groups to one of finding finite groups $K$ which admit a group of  automorphisms $H$ such that no non-identity element of $H$ fixes any non-identity element of $K.$ The subgroup $H$ is often known as a Frobenius complement, and the subgroup $K$ is known as a Frobenius kernel. A deep theorem of J.G. Thompson states that Frobenius kernels are nilpotent. It was known to W. Burnside that a Sylow $p$-subgroup 
of a Frobenius complement contains no elementary Abelian subgroup of order $p^{2}.$
    Perhaps the canonical example of a Frobenius group is the case that $K$ is the additive group of a finite field with $q$ elements, and $H$ is the multiplicative group of that field, which acts by multiplication (within the field) on $K$ (when $K$ is identified with the field).
 Hence for any prime power $q,$ there is a Frobenius group $G$ of order $q(q-1),$ your example $S_{3}$ being one manifestation of this.
  An example of a Frobenius group which is not solvable occurs with $H \cong {\rm SL}(2,5)$
and $K$ Elementary Abelian of order $121$, using the action of $H$ on $K$ given by the fact that $H$ is isomorphic to a subgroup of ${\rm SL}(2,11).$