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How do you prove Cayley's theorem which states that every group is isomorphic to a permutation group?
Wiki User
∙ 13y ago
Cayley's theorem:Let (G,$) be a group. For each g Є G, let Jg be a permutation of G such that
Jg(x) = g$x
J, then, is a function from g to Jg, J: g --> Jg and is an isomorphism from (G,$) onto a permutation group on G.
Proof:We already know, from another established theorem that I'm not going to prove here, that an element invertible for an associative composition is cancellable for that composition, therefore Jg is a permutation of G. Given another permutation, Jh = Jg, then h = h$x = Jh(x) = Jg(x) = g$x = g, meaning J is injective.Now for the fun part!
For every x Є G, a composition of two permutations is as follows:
(Jg â—‹ Jh)(x) = Jg(Jh(x)) = Jg(h$x) = g$(h$x) = (g$h)$x = Jg$h(x)
Therefore Jg ○ Jh = Jg$h(x) for all g, h Є G
That means that the set Ђ = {Jg: g Є G} is a stable subset of the permutation subset of G, written as ЖG, and J is an isomorphism from G onto Ђ. Consequently, Ђ is a group and therefore is a permutation group on G.
Q.E.D.
Wiki User
∙ 13y ago
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