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抄点不会的杨子胥抽代习题Day0

2023-02-17 21:00 作者:RIP_Official 0人读过 | 我要投稿

证明:设 $x%2Cy%5Cin%20G$ ,有 $%5Cvarphi(xy)%3D%5Cvarphi(x)%5Cvarphi(y)$ 和 $%5Cvarphi(yx)%3D%5Cvarphi(y)%5Cvarphi(x)$ .

有 $(xy)%5E3%3Dx%5E3y%5E3%0A$ 和 $(yx)%5E3%3Dy%5E3x%5E3$ ,

进而 $(yx)%5E2%3Dx%5E2y%5E2$ 和 $(xy)%5E2%3Dy%5E2x%5E2$ ,

同理 $(y%5E2x%5E2)%5E2%3Dx%5E4y%5E4%2C$

考虑 $(xy)%5E4%3D((xy)%5E2)%5E2%3D(y%5E2x%5E2)%5E2%3Dx%5E4y%5E4$ ，

从而 $(yx)%5E3%3Dx%5E3y%5E3$ ,即 $%5Cvarphi(xy)%3D%5Cvarphi(yx)$ ,由 $%5Cvarphi%0A$ 单,知道 $xy%3Dyx%0A$ ，即证.

证明:设 $%5Cvarphi%5Cin%20C(%7B%5Crm%20Aut%7DG)$ , $%5Cphi_a%3Ax%5Cto%20a%5E%7B-1%7Dxa%5Cin%5Crm%20InnG$ .

于是 $%5Cvarphi%5Cphi_a%3D%5Cphi_a%5Cvarphi$ ,即

$%5Cvarphi(a%5E%7B-1%7Dxa)%3Da%5E%7B-1%7D%5Cvarphi(x)%20a$ , $%5Cforall%20x%5Cin%20G$ .

这样 $%5Cvarphi(a)%5E%7B-1%7D%5Cvarphi(x)%5Cvarphi(a)%3Da%5E%7B-1%7D%5Cvarphi(x)a$ ,从而 $%5Cvarphi(a)a%5E%7B-1%7D%5Cvarphi(x)%3D%5Cvarphi(x)%5Cvarphi(a)a%5E%7B-1%7D$ ,

于是由 $%5Cvarphi(x)$ 任意性知道 $%5Cvarphi(a)%3Da$ ,由 $a$ 任意性知道 $%5Cvarphi%3D%5Crm%20Id$ .

第二个做的时候单纯地想到了InnG了，但是不知道和AutG搞在一起，完全不知道，总感觉有某些神奇手段避免和InnG接触，看了答案之后也不清楚为啥这样.这样不会那也忘了，寄了。

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