Question

State and prove Cayley's theorem.
Long qustion​

Answer 1

Answer:

In group theory, Cayley's theorem, named in honour of Arthur Cayley, states that every group G is isomorphic to a subgroup of the symmetric group acting on G.[1] This can be understood as an example of the group action of G on the elements of G.[2]

A permutation of a set G is any bijective function taking G onto G. The set of all permutations of G forms a group under function composition, called the symmetric group on G, and written as Sym(G).[3]

Cayley's theorem puts all groups on the same footing, by considering any group (including infinite groups such as (R,+)) as a permutation group of some underlying set. Thus, theorems that are true for subgroups of permutation groups are true for groups in general. Nevertheless, Alperin and Bell note that "in general the fact that finite groups are imbedded in symmetric groups has not influenced the methods used to study finite groups".[4]

The regular action used in the standard proof of Cayley's theorem does not produce the representation of G in a minimal-order permutation group. For example, {\displaystyle S_{3}} S_{3}, itself already a symmetric group of order 6, would be represented by the regular action as a subgroup of {\displaystyle S_{6}} S_{6} (a group of order 720).[5] The problem of finding an embedding of a group in a minimal-order symmetric group is rather more difficult.[6][7]