作者ol8889 (ol8889)
看板Math
标题[代数] Internal Direct Product 的疑问
时间Wed Jul 27 22:23:47 2011
根据Herstein 定义
A group is the internal direct product of normal subgroups N1,N2,...,Nm
if (a) G=N1 N2 ... Nm
(b) For any g of G , g=h1 h2 ... hm , hi 属於 Ni , uniquely.
我的疑问是
(1)G=N1 N2 ... Nm
and
(2)Ni∩Nj= (e) for i≠j (e=identity element)
可以imply (b) ?!
我的证法如下(请帮我看一下哪边出问题)
Proof:
for any g of G
put g=x1 x2 ... xm
=y1 y2 ... ym
(a′= inverse of a)
e=identiy element
=( g )(g′)
= 〔x1 x2 ... xm 〕〔 (ym)′ ... (y2)′ (y1)′〕
= 〔x1 x2 ...xm-1〕〔xm (ym)′〕〔(ym-1)′ ... (y2)′ (y1)′〕
(Since Nm-1∩Nm = (e) and Nm-1 and Nm are normal subgroups of G,
then 〔xm (ym)′〕(ym-1)′= (ym-1)′〔xm (ym)′〕 )
= 〔x1 x2 ...xm-1〕(ym-1)′〔xm (ym)′〕〔(ym-2)′... (y2)′ (y1)′〕
in the same way, we have
e=( g )(g-1)
=〔x1(y1)′〕〔x2(y2)′〕 ... 〔xm(ym)′〕 (X)
(如果懒得看我打的烂符号,
到这里就是
因为Ni∩Nj= (e) for i≠j, 後面的 (ym)′ ... (y2)′ (y1)′可以往前移 )
Claim: xi(yi)′=e for all i
Suppose xi(yi)′≠e for some i
by (X)
〔xi(yi)′〕′= xj(yj)′ for some j≠i ; otherwise (X) would not hold.
since〔xi(yi)′〕′= xj(yj)′ 属於 Nj,
xi(yi)′属於 Ni∩Nj=(e)
and thus xi = yi , contradiction.
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1F：→ wuxr :说真的有点忘了...但是〔xm (ym)′〕(ym-1)′= 07/28 14:13
2F：→ wuxr : (ym-1)′〔xm (ym)′〕..这里应该有问题. 07/28 14:14