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게시글 주소: https://orbi.kr/00039257454
Q: For a finite group $G$ of order $n>2$, $|Aut(G)|>1$.\\
Proof. Suppose $|Aut(G)|\leq 1$. $G$ acts on $G$ by conjugation. By the induced homomorphism, $G/Z(G)<Aut(G)$. Hence $|G/Z(G)| = 1$ i.e. $G$ is abelian. Note that the inversion map $x\mapsto x^{-1}$ is an automorphism. By assumption, $x = x^{-1}$ so that $x^2 =1$ for any $x\in G$. Hence, $|G| = 2^n$ where $n$ is the number of generators of $G$. Let $\{x_1,...,x_n\}$ be a generator of $G$. Then we can define a nontrivial automorphism $\varphi:G\to G$ by $x_1\mapsto x_2, x_2\mapsto x_1$ and $x_i\mapsto x_i$ for $3\leq i\leq n$ (Note that as $|G|>2$, $i\geq 2$) which is a contradiction. Hence, $|Aut(G)|>1$.
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