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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_{g}P_{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}.

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