What does regular and corefree mean (transitive permutation group)? I have to prove that a transitive permutation group, $G$, is regular.  What is the definition of regular?
In addition, my lecturer hinted that a transitive permutation group is regular if and only if there is no corefree proper subgroup.  My understanding is that the normal core of $H$ in $G$, with $H<G$, is the intersection of all the conjugates of $H$ (which is equivalent to several other definitions, http://groupprops.subwiki.org/wiki/Normal_core ).  From this, I've inferred that he means that for every subgroup, $H<G$, the core of $H$ in $G$ is the trivial group.  Is this the correct definition?
 A: A permutation group $G$ acting on $X$ (that is, a subgroup of $S_X$) is regular if and only if for ever $x\in X$, the stabilizer of $x$ is trivial; that is, the stabilizer of $x$,
$$G_x = \{g\in G\mid gx = x\},$$
equals the trivial group for every $x\in X$.
If $g\in G$ is such that $gx = y$, then it is an easy matter to verify that $G_x = gG_yg^{-1}$; in particular, if the action of $G$ is transitive, then $G$ is regular if and only if $G_x=\{1\}$ for at  least one (and hence for all) $x\in X$. 
Given a group $G$ and a subgroup $H$ of $G$, the core of $H$ is the largest normal subgroup of $G$ that is contained in $H$; as you note, this is equal to the intersection of all conjugates of $H$ in $G$. A subgroup $H$ of $G$ is core-free if the core of $H$ in $G$ is the trivial group.
Your inference that showing a subgroup is core-free is the same as showing that core is the trivial subgroup is correct, modulo a small caveat; corrected: However, the professor is saying you should show that there are no core-free proper subgroups, so you need to show that the core of any subgroup (other than the trivial group) is not the trivial group.
