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Let G be a group acting on a set X. The action is called faithful if for any g ≠ 1 ∈ G there exists an x ∈ X such that gx ≠ x. That is, only the identity fixes everything.
Prove the following:
(a) A group G acts faithfully on X if and only if the corresponding homomorphism Ψ: G -> AutSet(X) is injective. Thus, for a faithful action, G is isomorphic to a subgroup of AutSet(X).
(b) In general, G/ker Ψ acts faithfully on X. (You will need to dene how G= ker Ψ acts, and make sure this is an action.)
Let p, q and l be 3 distinct prime numbers, and let G be a finite group. Prove that G is solvable if:
2/2
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what is the answer using pemdas (32 divided into 4)+3
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