作者Royden (还名於世)
看板Math
标题Re: [代数] 代数(6)
时间Thu Oct 6 02:12:17 2005
※ 引述《TaiwanBank (PTT生日快乐^^)》之铭言：
: If A and B are subgroups of G with [G:A] < ∞, [G:B] < ∞, and
: [G:A] and [G:B] are relatively prime, then G = AB.
pf:
Let [G:A] = n [G:B] = m.
Let S := {aB | a \in A} and T := {gB | g \in G}.
Then t := |S| ≦ |T| = m.
Let S = {a_i B | i=1,2...,k } (i.e. a_i B = a_j B iff i=j)
Claim: A= ∪a_i(A∩B) (i.e. [A:A∩B]= k)
1.∪a_i(A∩B) is contained in A. (Trivial)
2.for a \in A
then aB= a_i B for some i
Hence a = a_i b for some b \in B
Since b = (a_i)^(-1) a \in A, b \in A∩B.
Therefore a \in ∪a_i B.
Now [G:A∩B]=[G:A][A:A∩B]=nt is finite
Similar [G:A∩B]=[G:B][B:A∩B]= mk is finite for some k.
Hence mk=nt. Since (m,n)=1, m|t. Hence m≦t.
But t ≦ m , therefore t=m. i.e. S=T.
G=AB.
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