Let G be a group and let a G Define a function fa G G by f

Let (G, ) be a group and let a G. Define a function f_a : G G by f_a(x) = a x. On the worksheet in class, we showed that the functions f_a are permutations. Let H be the set of such permutations H = {f_a : a G}. Prove that (G, ) = (H, ) where is the composition.

Solution

proof. Let G be a finite group. if a € G, then for every x in G the product ax is also an element of G. Now consider the function f_a from G into G defined by

f_a(x) = ax , for all x € G

The function f_a is one-one because if x , y € G ,then

f_a(x) = f_a(y) => ax = ay => x = y (by left cancellation law)

The function f_a is also onto because if x is any element of G , then there exists an element a^-1x in G such that

f_a(a^-1x) = a(a^-1x) = (aa^-1)x = ex = x

thus f_a is one -one function from G onto G. Therefore f_a is a permutation on G. Let H denote the set of all such one-one,onto functions defined on G corresponding to every element of G .

i.e H = {f_a : a € G }

ist we show that H is a group w.r.t the operation known as composite or product of two functions.

1.Closure property . let f_a, f_b € H ,where a, b € G. from our definitions of product of two functions , we have

(f_a f_b) (x) = f_a [f_b (x)] = f_a(bx) = a(bx) = (ab)x =f_ab(x) for all x € G

therefore by defn. of equality of 2 functions ,we have

f_a f_b = f_ab .......(1)

since ab € G , therefore f_ab € H ,and H is closed w.r.t product of functions

(ii) Associativity. Let f_{a ,} f_{b ,}f_c € H , where a,b , c € G, then

f_a(f_b f_c) = f_af _bc   (from 1 )

=f_{a(bc) ( from 1)}

   ₌f_{(ab)c (by associativity in G)}

₌ f_ab f_{c ( from 1)}

  = (f _af_b ) f_{c (}from 1 )

therfore the operation in H is associative.

(iii). Existence of identity . if e is the identity of G , then f_e is the identity of H ,because for every f_a in H ,we have

f_e fa = f_ea = f_a   and f _af_e = f_ae = f_a

_(iv) Existence of inverse . If a^-1 is the inverse of a in G ,the f_a^-1 is the inverse of f_a in H because

f_a^-1f_a = f_a^-1a = f_{e and} f_a f_a^-1 = f_aa^-1 = f_e

_{Thus H is a group}

Now we shall show that G ~= H. Consider the function Ø from G into H defined by Ø(a) = f_a ,for all a € G .

Ø is one -one ; if a,b € G , then

Ø(a) = Ø(b) which implies f_a= f_b => f_a(x) = f_b(x) for all x€ G

which implies ax = bx , for all x€ G which implies a=b,

therefore Ø is one-one

Ø is onto. let f_a be any element of H.

then for a € G and we have    Ø(a) = f_a, therefore   Ø is onto

Ø is homomorphism .if a,b € G ,then

Ø(ab) = f_ab

   =f_a f_{b =} Ø(a)   Ø(b)

thus   Ø is homomorphism

  therefore G is isomorphic to H.