When and how is a group of order n isomorphic to a regular subgroup of equal order? In "Group Theory and Its Application to Physical Problems" by Morton Hamermesh, Morton states Cayley's theorem: Every group G of order n is isomorphic with a subgroup of the symmetric group Sn, which makes sense to me.
Later the book discusses regular permutations and regular subgroups, and makes this statement: "...suppose that n is a prime number. Then the group of order n is isomorphic to a regular subgroup of Sn." (page 19 in the Dover edition)
Why is the last sentence true? Is every group of any order n isomorphic to a regular subgroup of Sn?
 A: A permutation group is called regular if it is transitive and all stabilizers are trivial.  Left multiplication yields a regular embedding of any group into its group of permutations (so the answer to your second question is "Yes").
A: To the OP: I think some of us were confused because "regular subgroup" is not a common thing to say if you're not a permutation group theorist.
As Scott points out, any abstract group can be thought of as acting regularly and transitively on some set (e.g. the set of group elements, by translation); so the question "is H a regular subgroup of G?" is only nontrivial if you've first specified a way in which G acts on some other set X  -- i.e. if G is given to you as a permutation group, not an abstract group.
(And yes, permutation group theorists appear to have many papers on "regular normal subgroups...")
A: As far as I can tell, "regular subgroup" means "transitive subgroup of order n of Sn," so I'm a bit confused by your question.  Are you confused about why it's transitive?  If that's the case, you need to read the proof of Cayley's theorem more carefully.
