│G│=6=2*3
by Chauchy's Theorem
there exists an element of order 3
let this element a
Because 3 is prime, it's trivial that A={e,a,a^2} is a subgroup of G
Assume there exists another subgroup B of order 3
Again because 3 is prime
so B is cyclic
let B={e,b,b^2}
if A=/=B
then consider A 交集 B
(一) if A 交集 B 不只有 e：
because│A 交集 B│會整除 │A│(by Lagrange's Theorem)
so │A 交集 B│= p
we have A=B , 矛盾
(二) if A 交集 B 只有 e：
i.e. a^i=/=b^j
consider AB = {a^i*b^j│a^i 屬於A , b^j 屬於B }
if a^i*b^j=a^i'*b^j'
then a^(i-i')=b^(j'-j)
then the only choose is a^(i-i')=b^(j'-j)=e
so AB has 3^2=9 distinct element, 矛盾
so A=B ------ subgroup of order 3 is unique
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補充：
because for all g 屬於 G
g^-1*A*g is a subgroup of G
and │g^-1*A*g│=3
so g^-1*A*g = A for all g 屬於 G
so A is normal in G
As a matter of fact
if │G│=pq, p>q, p,q are primes
then there exists a unique normal subgroup of order p in G
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◆ From: 111.251.230.217
※ 編輯: znmkhxrw 來自: 111.251.230.217 (12/31 00:23)