Let $G$ be a permutation group on $X= \{ x_1,x_2,...,x_n\}$

Prove that $Aut(X,G)=G$

I have no ideas how to prove it.

Besides, there is a question that the problem asks to prove $Aut(X,G) = G $ , i.e. prove automorphism group with the structure (permutation group G) is the permutation group.

But, in context the textbook says that we indeed prove every permutation group is the automorphism group of some object .

What's the connected?

Here are some definitions from textbook

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By definition $\operatorname{Aut}(X,G)$ is a subset of $S(X)$, the group of all permutations of $X$. By definition a permutation $\sigma\in S(X)$ is an automorphism of $(X,G)$ if and only if $G\sigma=G$. In particular this implies that $e\sigma=\sigma\in G$, which shows that $\operatorname{Aut}(X,G)\subset G$. On the other hand, for all $g\in G$ you have $Gg=G$ and hence $g\in\operatorname{Aut}(X,G)$, which shows that $G\subset\operatorname{Aut}(X,G)$. Hence $\operatorname{Aut}(X,G)=G$.

This shows that $G$ is the automorphism group of $(X,G)$. As $G$ was an arbitrary permutation group to begin with, this proves that every permutation group is an automorphism group.

  • $\begingroup$ Thanks! I have realized almost things but the $e\sigma=\sigma\in G$. The $\sigma$ is permutation like $(g_1) \sigma = g_2$. How can I derive it? $\endgroup$ – matchz Dec 12 '18 at 18:31
  • $\begingroup$ Well $g_1,g_2\in G$ and $G$ is a group, so $g_1^{-1}g_2\in G$. $\endgroup$ – Servaes Dec 12 '18 at 18:37
  • $\begingroup$ Or $\sigma$ is not a permutation function ($S(X)$)? how to say a function is a element in $G$ $\endgroup$ – matchz Dec 12 '18 at 18:58
  • $\begingroup$ By definition $G$ is a subset of $S(X)$. $\endgroup$ – Servaes Dec 12 '18 at 20:24

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