# Cayley's theorem

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Cayley's theorem

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

A permutation of a set "G" is any bijective function taking "G" onto "G"; and the set of all such functions forms a group under function composition, called "the symmetric group on" "G", and written as Sym("G").

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 which are true for permutation groups are true for groups in general.

History

Attributed in Burnside [Citation | last = Burnside | first = William | author-link = William Burnside | title = Theory of Groups of Finite Order | location = Cambridge | year = 1911 | edition = 2] to Jordan [Citation | last = Jordan | first = Camille | author-link = Camille Jordan | title = Traite des substitutions et des equations algebriques | publisher = Gauther-Villars | location = Paris | year = 1870] , Eric Nummela [Citation | last = Nummela | first = Eric | title = Cayley's Theorem for Topological Groups | journal = American Mathematical Monthly | volume = 87 | issue = 3 | year = 1980 | pages = 202-203] nonetheless argues that the standard name for this theorem -- "Cayley's Theorem" -- is in fact appropriate. Cayley, in his original 1854 paper [Citation | last = Cayley | first = Arthur | author-link = Arthur Cayley | title = On the theory of groups as depending on the symbolic equation θn=1 | journal = Phil. Mag. | volume = 7 | issue = 4 | pages = 40-47 | year = 1854] which introduced the concept of a group, showed that the correspondence in the theorem is one-to-one, but he failed to explicitly show it was a homomorphism (and thus an isomorphism). However, Nummela notes that Cayley made this result known to the mathematical community at the time, thus predating Jordan by 16 years or so.

Proof of the theorem

From elementary group theory, we can see that for any element "g" in "G", we must have "g"*"G" = "G"; and by cancellation rules, that "g"*"x" = "g"*"y" if and only if "x" = "y". So multiplication by "g" acts as a bijective function "f""g" : "G" → "G", by defining "f""g"("x") = "g"*"x". Thus, "f""g" is a permutation of "G", and so is a member of Sym("G").

The subset "K" of Sym("G") defined as "K" = {"f""g" : "g" in "G" and "f""g"("x") = "g"*"x" for all "x" in "G"} is a subgroup of Sym("G") which is isomorphic to "G". The fastest way to establish this is to consider the function "T" : "G" → Sym("G") with "T"("g") = "f""g" for every "g" in "G". "T" is a group homomorphism because (using "•" for composition in Sym("G")):("f""g" • "f""h")("x") = "f""g"("f""h"("x")) = "f""g"("h"*"x") = "g"*("h"*"x") = ("g"*"h")*"x" = "f"("g"*"h")("x"), for all "x" in "G", and hence: "T"("g") • "T"("h") = "f""g" • "f""h" = "f"("g"*"h") = "T"("g"*"h"). The homomorphism "T" is also injective since "T"("g") = id"G" (the identity element of Sym("G")) implies that "g*x" = "x" for all "x" in "G", and taking "x" to be the identity element "e" of "G" yields "g" = "g"*"e" = "e". Alternatively, "T"("g") is also injective since, if "g"*"x"="g"*"x' " implies "x"="x' " (by pre-multiplying with the inverse of "g", which exists because "G" is a group).

Thus "G" is isomorphic to the image of "T", which is the subgroup "K".

"T" is sometimes called the "regular representation of" "G".

Alternate setting of proof

An alternate setting uses the language of group actions. We consider the group $G$ as a G-set, which can be shown to have permutation representation, say $phi$.

Firstly, suppose $G=G/H$ with $H=\left\{e\right\}$. Then the group action is $g.e$ by classification of G-orbits (also known as the orbit-stabilizer theorem).

Now, the representation is faithful if $phi$ is injective, that is, if the kernel of $phi$ is trivial. Suppose $g$ ∈ ker $phi$ Then, $g=g.e=phi\left(g\right).e$ by the equivalence of the permutation representation and the group action. But since $g$ ∈ ker $phi$, $phi\left(g\right)=e$ and thus ker $phi$ is trivial. Then im $phi < G$ and thus the result follows by use of the first isomorphism theorem.

Remarks on the regular group representation

The identity group element corresponds to the identity permutation. All other group elements correspond to a permutation that does not leave any element unchanged. Since this also applies for powers of a group element, lower than the order of that element, each element corresponds to a permutation which consists of cycles which are of the same length: this length is the order of that element. The elements in each cycle form a left coset of the subgroup generated by the element.

Examples of the regular group representation

Z2 = {0,1} with addition modulo 2; group element 0 corresponds to the identity permutation e, group element 1 to permutation (12).

Z3 = {0,1,2} with addition modulo 3; group element 0 corresponds to the identity permutation e, group element 1 to permutation (123), and group element 2 to permutation (132). E.g. 1 + 1 = 2 corresponds to (123)(123)=(132).

Z4 = {0,1,2,3} with addition modulo 4; the elements correspond to e, (1234), (13)(24), (1432).

The elements of Klein four-group {e, a, b, c} correspond to e, (12)(34), (13)(24), and (14)(23).

S3 (dihedral group of order 6) is the group of all permutations of 3 objects, but also a permutation group of the 6 group elements:

*Yoneda lemma

References

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