Discussion
Let G be a group and a map \(Q : G \to G\) defined by \(Q(a) = a^{-1}\). Then which of the following is false ?
Answer: D
The false statement is Q is a homomorphism if G is any permutation group.
Inversion satisfies \(Q(ab) = (ab)^{-1} = b^{-1}a^{-1}\), while \(Q(a)Q(b) = a^{-1}b^{-1}\). These agree for all a, b exactly when \(ab = ba\) for all a, b. So Q is a homomorphism if and only if G is abelian, which makes the second and third statements true. Q is always bijective, being its own inverse.
A permutation group such as \(S_{3}\) is not abelian: with \(a = (1\,2)\) and \(b = (1\,3)\), \((ab)^{-1} \neq a^{-1}b^{-1}\). So Q is not a homomorphism for every permutation group.
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