看板 Math 關於我們 聯絡資訊
方法大同小異 僅供參考- Let F be a free abelian group. We can find a basis {e_i}, i is in an index set I. Then F = Σ <e_i>. i in I Let j be in I. Let A = Σ Ci, Ci = <e_i> if i≠j, Cj = <n*e_j>. i in I (Where n*e_j = e_j +...+ e_j (n-times)) Clearly, A is a subgroup of F and F/A is isomorphic to Σ Si, Si = <e_i>/<e_i> if i≠j, Sj = <e_j>/<n*e_j> i in I Then F/A is isomorphic to Zn since <e_j>/<n*e_j> is isomorphic to Zn, hence, [F:A] = n. ※ 引述《hotplushot (熱加熱)》之銘言: : ※ 引述《hotplushot (熱加熱)》之銘言: : : A nonzero free abelian group has a subgroup of index n : : for all positive integer n. : : 這題毫無頭緒..... : : 可否請版友協助 : : 感謝!!!! : because F is free abelian : F=Z⊕...⊕Z(m個) : let A=Z⊕...⊕nZ⊕...⊕Z : [F:A]=n : and nZ is subgroup of Z : so conclusion is hold. : 是這樣嗎 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 111.251.166.99
sato186 :啊 被搶PO了 03/18 02:51
VFresh :so? 03/18 05:59
VFresh :你高興可以再PO一篇 我這篇砍掉 03/18 06:00