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Proposition 2. Goφ G′ defined above is a group. Proof. (1). For any (g1, g ′ 1), (g2, g ′ 2), (g3, g ′ 3) ∈ Goφ G′, ((g1, g ′ 1)(g2, g ′ 2))(g3, g ′ 3) = (g1φg′ 1(g2), g ′ 1g ′ 2)(g3, g ′ 3) = (g1φg′(More)