Re: [代數] 自由群的問題
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
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