Re: [代數] 代數(5)
※ 引述《PttFund (批踢踢基金只進不出)》之銘言：
: Let G be an abelian group of order 360.
: (a) How many Sylow 3-subgroups does G have?
Because G is abelian, every subgroup of G is normal in G.
Hence G has only one Sylow 3-subgroup.
: (b) Up to isomorphism, describe all the possible structures of G.
By fundamental theorem of finite abelian groups, we have
G = Z_360 ps: Z_m x Z_n = Z_mn if and only if (m,n)=1
= Z_180 x Z_2
.
.
.
(依此類推)
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◆ From: 140.122.140.213
※ 編輯: elvolvo 來自: 140.122.140.213 (07/24 16:59)
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