Let G be a group, and fix an element g ∈ G. Define a function Ag : G → G by Ag(x) = gx. Prove that for all g ∈ G, Ag is a permutation of G. Also, prove that G is abelian if and only if Ag is the identity function on G for all g ∈ G.
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Step 1
To do this, we need to show that Ag is both injective and surjective. Injectivity: Suppose Ag(x) = Ag(y) for some x, y € G. Then we have (x, gx) = (y, gy), which implies x = y since g is fixed. Therefore, Ag is injective. Surjectivity: Let (a, b) € G x | 811 xg. Show more…
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