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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 $Hhttp://groupprops.subwiki.org/wiki/Normal_core ). From this, I've inferred that he means that for every subgroup, $H

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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 $H$H\leq G$2011-11-26
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    @Zeophlite: The notation `<` for subgroups is not always read as "proper"; overkill is certainly better than leaving it unclear.2011-11-26
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    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 $G$.2011-11-26
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    @Derek: Yes, the question is somewhat strangely phrased from the reports. [See here](http://math.stackexchange.com/questions/85714/what-does-this-group-permute)2011-11-26
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    @Jack: How so? A subgroup if core-free if the core is the trivial group, isn't it?2011-11-26
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    @Jack: Ah, yes; the "hint" by the professor, as opposed to the meaning of "core-free". Thanks.2011-11-26
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    Looks good! $ $2011-11-26