# Cayley's theorem

In group theory, Cayley's theorem, named in honour of Arthur Cayley, states that every finite group G is isomorphic to a subgroup of the symmetric group acting 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. The set of all permutations of G 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 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".

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, $S_{3}$ , itself already a symmetric group of order 6, would be represented by the regular action as a subgroup of $S_{6}$ (a group of order 720). The problem of finding an embedding of a group in a minimal-order symmetric group is rather difficult.

## History

While it seems elementary enough, at the time the modern definitions didn't exist, and when Cayley introduced what are now called groups it wasn't immediately clear that this was equivalent to the previously known groups, which are now called permutation groups. Cayley's theorem unifies the two.

Although Burnside attributes the theorem to Jordan, Eric Nummela nonetheless argues that the standard name—"Cayley's Theorem"—is in fact appropriate. Cayley, in his original 1854 paper, showed that the correspondence in the theorem is one-to-one, but he failed to explicitly show it was a homomorphism (and thus an embedding). However, Nummela notes that Cayley made this result known to the mathematical community at the time, thus predating Jordan by 16 years or so.

The theorem was later published by Walther Dyck in 1882 and is attributed to Dyck in the first edition of Burnside's book.

## Proof of the theorem

If g is any element of a group G with operation ∗, consider the function fg : GG, defined by fg(x) = gx. By the existence of inverses, this function has a two-sided inverse, $f_{g^{-1}}$ . So multiplication by g acts as a bijective function. Thus, fg is a permutation of G, and so is a member of Sym(G).

The set K = {fg : gG} is a subgroup of Sym(G) that is isomorphic to G. The fastest way to establish this is to consider the function T : G → Sym(G) with T(g) = fg for every g in G. T is a group homomorphism because (using · to denote composition in Sym(G)):

$(f_{g}\cdot f_{h})(x)=f_{g}(f_{h}(x))=f_{g}(h*x)=g*(h*x)=(g*h)*x=f_{g*h}(x),$ for all x in G, and hence:

$T(g)\cdot T(h)=f_{g}\cdot f_{h}=f_{g*h}=T(g*h).$ The homomorphism T is injective since T(g) = idG (the identity element of Sym(G)) implies that gx = x for all x in G, and taking x to be the identity element e of G yields g = ge = e, i.e. the kernel is trivial. Alternatively, T is also injective since gx = g′ ∗ x implies that g = g (because every group is cancellative).

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

T is sometimes called the regular representation of G.

### Alternative setting of proof

An alternative 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 $H=\{e\}$ . 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 $g\in \ker \phi$ Then, $g=g.e=\phi (g).e$ by the equivalence of the permutation representation and the group action. But since $g\in \ker \phi$ , $\phi (g)=e$ and thus $\ker \phi$ is trivial. Then $\mathrm {Im} \,\phi \cong G$ and thus the result follows by use of the first isomorphism theorem.

## Remarks on the regular group representation

The identity element of the group corresponds to the identity permutation. All other group elements correspond to derangements: permutations that do 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 that consists of cycles all of the same length: this length is the order of that element. The elements in each cycle form a right 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). E.g. 0 +1 = 1 and 1+1 = 0, so 1 -> 0 and 0 -> 1, as they would under a permutation.

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, and the latter is how it is realized by its regular representation.

* e a b c d f permutation
e e a b c d f e
a a e d f b c (12)(35)(46)
b b f e d c a (13)(26)(45)
c c d f e a b (14)(25)(36)
d d c a b f e (156)(243)
f f b c a e d (165)(234)

## More general statement of the theorem

A more general statement of Cayley's theorem consist of considering the core of an arbitrary group $G$ . In general if $G$ is a group and $H$ is a subgroup with $|G:H|<\infty$ , then $G/{\text{Core}}_{G}(H)$ is isomorphic to a subgroup of $\Sigma _{|G:H|}$ . In particular if $G$ is a finite group and we set $H=1$ then we get the classic result.