Cayley’s theorem | Algebra - Mathematics PDF Download

We have stated that on of the main objectives of group theory is to write down a complete list of non-isomorphic groups. At first, such a task appears hopeless. For, as we have seen, groups pop up in some very unexpected places and, therefore, if we set out to compile a list of all non-isomorphic groups, we would hardly begin to know where to look. The following theorem of Cayley solves this dilemma.

Theorem 1: Every group is isomorphic to a subgroup of a permutation group

Proof: Let G be a group, g  G. define

If P_g(x) = P_g(y), then xg^-1 = yg^-1, so that x = y. Therefore, P_g is an injection. If y  G, then P_g(yg) = yg · g^-1 = y. Therefore, P_g is a surjection. Thus, since P_g is a bijection, P_g is a permutation of the elements of the set G, and P_g ∈ S_G. Let us consider the mapping

G → S_G

defined by

g  P_g                           (1)

Since P_gg' = x(gg')^-1 = P_gP_g'(x), the mapping (1) is a homomorphism. But P = 1_SG if and only if x · g = x for all x  G, which occurs if and only if g = 1_G. Therefore, the kernel of the homomorphism (1) is 1_G, and therefore the mapping (1) is an injection. Thus we have shown that G is isomorphic to a subgroup of S_G.

What Cayley's theorem tells us is that permutation groups and their subgroups are all the groups that can exist. Unfortunately, the problem of classifying the subgroups of a permutation group is extremely complicated, even in the case of a finite permutation group. Therefore, Cayley's theorem does not allow us to easily identify a complete list of groups.

The above argument actually proves somewhat more than claimed. For if G is finite, having order n, then G is isomorphic to a subgroup of S_G. Therefore we have

Corollary 2: If G has finite order n, then G is isomorphic to a subgroup of S_n.