Re: [代數]
※ 引述《k591 (焦。糖。瓶。果)》之銘言：
: 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.
: 拜託給個方向!
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◆ From: 219.71.210.134
※ 編輯: yusd24 來自: 219.71.210.134 (01/06 19:24)
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01/06 19:25, , 1F
01/06 19:25, 1F
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