|Aut(G)|=|G|

**petter** · January 27th 2009, 04:58 AM

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 .

**NonCommAlg** · January 27th 2009, 03:26 PM

Originally Posted by petter

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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