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Math Help - |Aut(G)|=|G|

  1. #1
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    |Aut(G)|=|G|

    Let G be a abelian group with |G|=|End(G)|=n . Prove that G are isomorphic with Z_{n}


    End(G) ={f:G->G with f(xy)=f(x)f(y) } . If G is abelian then End(G) is group .
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  2. #2
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    Quote Originally Posted by petter View Post
    Let G be a abelian group with |G|=|End(G)|=n . Prove that G are isomorphic with Z_{n}


    End(G) ={f:G->G with f(xy)=f(x)f(y) } . If G is abelian then End(G) is group .
    i think you should change the title of your post to "|End(G)|=|G|", because right now it has nothing to do with your post! anyway, this is a good question, especially for those who would like to

    see non-trivial applications of fundamental theorem of finite abelian groups. to solve the problem, i'll assume that G is not cyclic and i'll show that |\text{End}(G)|> |G|=n:

    we have G = <x_1> \oplus <x_2> \oplus \cdots \oplus <x_k>, where o(x_j)=n_j, \ 1 \leq j \leq k, and n_k \mid n_{k-1} \mid \cdots \mid n_2 \mid n_1. let A=\{(i_1,i_2, \cdots , i_k): \ \ 1 \leq i_1 \leq n_1, \ 1 \leq i_2 \leq n_2, \cdots, 1 \leq i_k \leq n_k \}. for any

    \alpha=(i_1,i_2, \cdots , i_k) \in A define f_{\alpha}: G \longrightarrow G by: f_{\alpha}(x_1^{r_1}x_2^{r_2} \cdots x_k^{r_k})=x_1^{r_1 i_1}x_2^{r_2i_2} \cdots x_k^{r_ki_k}. it's easy to see that f_{\alpha} \in \text{End}(G) and f_{\alpha} = f_{\beta} if and only if \alpha=\beta.

    therefore: |\text{End}(G)| \geq |\{f_{\alpha}: \ \alpha \in A \}|=|A|=|G|. the only thing left to prove is that: \exists \ g \in \text{End}(G): \ g \notin \{f_{\alpha}: \ \alpha \in A \}. so we define g: G \longrightarrow G by: g(x_1^{r_1}x_2^{r_2} \cdots x_k^{r_k})=x_2^{r_1}. note that g

    is well-defined because n_2 \mid n_1. it's clear that g \in \text{End}(G). now suppose that g=f_{\alpha}, for some \alpha \in A. then x_2=g(x_1)=f_{\alpha}(x_1)=x_1^i, for some 1 \leq i \leq n_1, which is obviously nonsense! \Box
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