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Iff:G->Hisahomomorphism,HisabelianandNisasubgroupofGcontainingKerf,thenNisnormalinG.GTM73里的一道题目,和群论有关1chenlei111222:f不是同构(isomorphism),而是同态(homomorphism)。换句话说它可能不是一
题目详情
If f:G -> H is a homomorphism,H is abelian and N is a subgroup of G containing Ker f,then N is normal in G.
GTM73里的一道题目,和群论有关
1
chenlei111222:
f不是同构(isomorphism),而是同态(homomorphism)。换句话说它可能不是一一的,只是一个群映射产生的对应。这样f(ab)= f(a)f(b)=f(ba)完全推不出ab=ba,所以您的推理是错误的。
您的翻译很准确。
2
524341170,你的回答是在胡说八道
GTM73里的一道题目,和群论有关
1
chenlei111222:
f不是同构(isomorphism),而是同态(homomorphism)。换句话说它可能不是一一的,只是一个群映射产生的对应。这样f(ab)= f(a)f(b)=f(ba)完全推不出ab=ba,所以您的推理是错误的。
您的翻译很准确。
2
524341170,你的回答是在胡说八道
▼优质解答
答案和解析
取N中任意元素n, G中任意元素g, 定义m=g^(-1)ng则
f(m)=f(g^(-1)ng)=f(g^(-1))f(n)f(g)=f(g)^(-1)f(n)f(g)=f(g)^(-1)f(g)f(n)=f(n).
所以
f(mn^(-1))=e
mn^(-1) is in Kerf, and further more in N.
Obviously n^(-1) is in N also.
Thus m is in N.
f(m)=f(g^(-1)ng)=f(g^(-1))f(n)f(g)=f(g)^(-1)f(n)f(g)=f(g)^(-1)f(g)f(n)=f(n).
所以
f(mn^(-1))=e
mn^(-1) is in Kerf, and further more in N.
Obviously n^(-1) is in N also.
Thus m is in N.
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