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Let G be a finite group and let X be a G-set. Assume that |G| does not divide |X|. Prove that there exists a gEG and an EX such that ge and such that gx = x. Hint: Use Burnside's formula.

User Couto
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Final answer:

Using Burnside's lemma, one can prove there exists a non-identity element g in G and an element x in G-set X such that gx = x by showing that the action of G on X cannot have all orbits of size equal to |G| since |G| does not divide |X|.

Step-by-step explanation:

The student is dealing with a concept in group theory, specifically with Burnside's lemma (or Burnside's formula), which is a result in group theory and combinatorics. To prove that there exists an element g in group G and an element x in G-set X such that g is not equal to the identity element and gx = x, we can use Burnside's lemma, which relates the sizes of a group and a set on which it acts. Burnside's lemma states that the average number of points fixed by elements of G equals the number of orbits under the action of the group. If the order of G does not divide the order of X, then the action of G on X cannot have all orbits of size equal to |G|. Hence, some points must be fixed by a group element that is not the identity, implying the existence of such g and x. This mathematical proof process illustrates the application of Burnside's formula in answering questions about group actions.

User McUsr
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