※ 引述《loribank (小蘿莉銀行)》之銘言：
: 有一題證明想了很久還是沒有方向
: 希望代數強者教我一下
: Suppose G is a group of order 6
: Prove：There is exactly one subgroup of order 3.
: 拜託了.....感謝!!
缺錢 xd
I want to skip Cauchy's and Sylow's them as the group order is too small.
If every element of G satisfies g^2=1, then G is abelian and then G=Z/6
is cyclic. The result is hence.
If not, there is some g in G s.t. o(g)>2.
By Lagrang them, o(g)=3 or 6. If o(g)=6 then G is cyclic, the result follows.
Now we suppose o(g)=3.
Take any elements x in G - {1,g,g^2}.
If o(x)=3, note that <x> (and <g>) is normal in G ,and so |<x><g>|=9>6
which is a contradiction.
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※ 編輯: Sfly 來自: 131.215.6.92 (12/31 06:32)