Let graph $G$ be isomorphic with $H$. I would like to show $\operatorname{Aut}(G)=\operatorname{Aut}(H)$, where $\operatorname{Aut}(G)$=Set of automorphisms of graph $G$).
Graph isomorphic
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graph-theory
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0The *sets* of automorphisms are hardly ever the *same*. But something stronger is true. The automorphism *groups* are isomorphic. My feeling is that the assertion does not even require proof. But if one wants a proof, it is automatic. – 2011-12-22
1 Answers
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HINT: If $G$ is isomorphic with $H$, there exists an isomorphism $\varphi$ from $G$ to $H$. Now, let $h \in \operatorname{Aut}(G)$, then $\varphi\circ h \circ \varphi^{-1} \in \operatorname{Aut}(H)$. Also, for each $h$ this automorphism is unique, thus $|\operatorname{Aut}(G)| \leq |\operatorname{Aut}(H)|$. Can you take it from here?
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0if $h \in Aut(G)$ and $\varphi^{-1} \in Aut(H)$implies $\varphi\circ h \circ \varphi^{-1} \in Aut(H)$why? – 2011-12-22
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0$\varphi^{-1}$ is not a member of $Aut(H)$, however, $\varphi^{-1}$ is an isomorphism from $H$ to $G$ and composition of two isomorphisms is again an isomorphism, so $\varphi\circ h \circ \varphi^{-1}$ is an isomorphism. Do you see that $\varphi\circ h \circ \varphi^{-1}$ is a function from $H$ to $H$? – 2011-12-22
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0yes,thanks Dimitri your proof is really impressing – 2011-12-22
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0Thanks, no problem, consider accepting it if you like it that much:). – 2011-12-22
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0Well, my hint shows that $|Aut(G)|\leq |Aut(H)|$. Can you now show yourself that $|Aut(H)| \leq |Aut(G)|$ (the method is analogous)? Also, i don't directly prove that $Aut(G) \cong Aut(H)$. This however also implies that they have the same size though (see http://en.wikipedia.org/wiki/Cardinality). – 2011-12-22
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0define $S:Aut(G) \mapsto Aut(H)$ such that $S(g)=\varphi\circ g \circ \varphi^{-1}$ then S is group Automorphism but I looking for $\operatorname{Aut}(G)=\operatorname{Aut}(H)$ – 2011-12-22
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0Well, you said that you could not show that $|Aut(G)| = |Aut(H)|$, and the proof for that is what I gave you. – 2011-12-22
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0To be honest, i don't think that both groups (unless H = G) can be equal to one another, since the domains and codomains of the automorphisms are different, thus the functions cannot be the same. However, they can be isomorphic as you already mentioned. – 2011-12-22