# 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?

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What does "regular" mean? What does "transitive subgroup" mean? Does it mean that the induced action of the subgroup on the n element set is transitive? –  Anton Geraschenko Oct 12 '09 at 16:15
A group of permutations of a nonempty set is transitive if the set has one orbit, i.e., if for any pair of elements of the set, there is an element of the group taking one to the other. –  S. Carnahan Oct 12 '09 at 17:10

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").

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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.

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