In the general case (where (G,*) can be any group) assume x 1 and x 2 are in G and y ∈ A.a) Use the definition to simplify ( π x 1 , π x 2 ) ( y ) .b) Use the definition to simplify π x 1 ⋅ x 2 ( y )If you've done this correctly the two answers should be equal. This shows that the two functions π x 1 ⋅ π x 2 and π x 1 ⋅ x 2 are equal.

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In the general case (where (G,*) can be any group) assume       x    1   and       x    2   are in G and   y  ∈  A.a) Use the definition to simplify   (      π          x      1        ,      π                  x        2              )  (  y  )  .b) Use the definition to simplify       π          x      1      ⋅      x      2        (  y  )If you&#039;ve done this correctly the two answers should be equal. This shows that the two functions       π          x      1        ⋅      π          x      2         and       π          x      1      ⋅      x      2       are equal.

Skyler Carney · Accepted Answer

Here       π                  x        1              (  y  )  =      x    1    ⋅  yNow   (      π          x      1        ⋅      π          x      2        )  (  y  )  =      π          x      1        (      π          x      2        (  y  )  )  =      π          x      1        (      x    2    ⋅  y  )  =      x    1    ⋅  (      x    2    ⋅  y  )  =  (      x    1    ⋅      x    2    )  ⋅  y (since associativity holds)but      π          x      1      ⋅      x      2        (  y  )  =  (      x    1    ⋅      x    2    )  ⋅  y by the defenition.Cayley&#039;s theorem states that every group G is isomorphic to a subgroup of the symmetric group of G (symmetric group of G is the set off all permutations on G). For each g belong to G we form πg and then form the subgroup of the symmetric group of G that is isomorphic to it, by picking together all these πgs. Here, we talk about a subgroup A of G and then also such a formation is valid since A itself is a group.

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