※ 引述《k591 (焦。糖。瓶。果)》之铭言： : 1.If G is nonabelian of order p^3 (p is a prime) : (a) |Z(G)| (hint: Z(G) > {1}) The center of a p-group G, Z(G), is always non-trivial. So |Z(G)| = p, p^2, or p^3. Case 1, |Z(G)|=p^3, G is abelian. Contradiction. Case 2, |Z(G)|=p^2, G/Z(G) is cyclic. So G is abelian. Contradiction. These implies |Z(G)|=p. : (b) Show that G'=Z(G) Note that G/Z(G) is abelian (Group of order p^2) => Z(G) contains G'. But the only subgroup of Z(G) is {1} or Z(G). And {1}≠G'(G/{1} is non-abelian) So G'=Z(G). : 2.Show that Inn(A5) isomorphism to A5 Consider A_5 → Inn(A_5) by g → f_g, where f_g(x)=gxg^{-1}. This is surjective by definition, This is injective since A_5 is simple. So it's an isomorphism. : 拜托给个方向! --

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