On the set K₃(G) of finite groups elements commuting exactly with three elements

Let G be an arbitrary finite multiplicative group, |G| = n. We define the set K₃(G) as follows: K3(G) = {x e G | |C_g(x)| = 3} = {x e G | Cg(x) = {e, x, x²}}. It follows from the definition of K₃(G) that A) if х e K₃(G), then the order of x is 3 (о(х) = 3); B) if х e K₃(G), then х² e K₃(G). The following properties of the set K₃(G) have been proved. Proposition 1. If K₃(G) * 0, then |G| : 3 and |G| / 9. Proposition 2. If х e K₃(G), then x^g e K₃(G) for each g e G. Proposition 3. Let K₃(G) * 0, х e G and o(x) = 3. Then х e K₃(G). Proposition 4. Let |G| = n; K₃(G) * 0. Then |K₃(G)| e {{; у Lemma 5. Let a, g e G, o(a) = 3; g~'ag = a². Then o(g):2 . Proposition 6. 1) Let o(a) = 3 and g^ag = a². Then | G | :6 . 2) If |G| = 2k + 1, then K₃(G) = 0 or |K₃ (G)| = ^Gi. Theorem 7. Let G be a finite simple group, |G| = n, K₃(G) * 0. Then all involutions of the group G form a class of conjugate elements.

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