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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, ). 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?

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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.

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Is proper necessary? I wrote $H<G$, not $H\leq G$ – Zeophlite Nov 26 '11 at 5:23
@Zeophlite: The notation < for subgroups is not always read as "proper"; overkill is certainly better than leaving it unclear. – Arturo Magidin Nov 26 '11 at 5:24
I am confused by this discussion! Every group can be represented as a regular permutation group, irrespective of whether its subgroups are core-free or not. What is true is that there being no nontrivial core-free subgroup of $G$ is equivalent to the regular representation being the only faithful transitive permutation representation of $G$. – Derek Holt Nov 26 '11 at 12:49
@Derek: Yes, the question is somewhat strangely phrased from the reports. See here – Arturo Magidin Nov 26 '11 at 20:48
@Jack: How so? A subgroup if core-free if the core is the trivial group, isn't it? – Arturo Magidin Nov 26 '11 at 20:49

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