Re: [代數] 代數(1)
※ 引述《PttFund (批踢踢基金只進不出)》之銘言：
: Let G be a group and let Z = Z(G) be its center. Show that
: if G/Z is cyclic, then G is abelian.
let xZ generates G/Z, then
any element of G has the form x^ky, where y belongs to Z
now (x^ky)(x^k'y')=x^(k+k')(yy')=(x^k'y')(x^ky)
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