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Cayley Theorem. Every group is isomorphic to a permutation group. Example: U(10). U(10) = {1, 3, 7, 9} Definition: For g in U(10), let T g (x)= gx T 1 (x) = T 3 (x) = T 7 (x) = T 9 (x) =. x. T 1 =. . (1 3 9 7). 3x. T 3 =. (1 7 9 3). 7x. T 7 =. 9x. T 9 =.
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Cayley Theorem Every group is isomorphic to a permutation group.
Example: U(10) • U(10) = {1, 3, 7, 9} • Definition: For g in U(10), let Tg(x)= gx • T1(x) = • T3(x) = • T7(x) = • T9(x) = x T1 = (1 3 9 7) 3x T3 = (1 7 9 3) 7x T7 = 9x T9 = (1 9)(3 7)
Every group is isomorphic to a permutation group Proof: Let G be any group. • For g in G, define Tg(x) = gx. We show Tg is a permutation on G. • Let S = {Tg | for g in G} We show S is a permutation group. • Define the map :G S by (g)=Tg We show is an isomorphism.
1. Tg is a permutation on G. Suppose Tg(x) = Tg(y). Then gx = gy. By left cancellation, x=y. Hence Tg is 1 to 1. Choose any y in G. Let x = g-1y Then Tg(x) = gx = gg-1y = y So Tg is onto. This shows that Tg is a permutation.
2. {Tg | g in G} is a group The operation is composition. For a,b,x, TaTb(x) = Ta(bx) = a(bx) = (ab)x =Tab(x) So TaTb = Tab (*) From (*), TeTa = Tea = Ta, So Te is the identity in S. If b = a-1 we have, TaTb = Tab= Te So Ta-1 = Tb and S has inverses. Function composition is associative. Therefore, S is a group.
3. (g) = Tg is isomorphism • Choose a, b in G. Suppose (a) = (b). Then Ta = Tb. In particular, for any x in G, Ta(x) = Tb(x) ax = bx a = b Therefore is one-to-one.
(g) = Tg is isomorphism • Choose any Tg in S. Then (g) = Tg Therefore, is onto.
(g) = Tg is isomorphism • Choose any a, b in G. Then (ab) = Tab = TaTb by (*) = (a) (b) Therefore, is Operation Preserving. It follows that is an isomorphism.
Why do we care? • The permutation group we constructed is called the Left Regular Representation of G. • Every abstract group can be represented in a concrete way • It shows that abstract groups are all permutation groups, unifying the study of both.
