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let $A$ be a non empty set with a given structure, and let $G$ be the group of automorphisms of $A$, that is, the set of bijective maps from $A$ to itself that preserve its structure, with composition as group law.

Is it true that for any $a\in A$, the subgroup $G(a)$ of $G$ consisting of all elements of $G$ that preserve $a$ is a normal subgroup of $G$? Thanks in advance.

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    As you can see below, the answer is no. As an exercise, you can try to prove (if you haven't done this already) that if two elements lie in the same orbit, then their stabilisers are conjugate. This is the best "normality" result you can get.2012-08-15

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No. Let $A = \{1,2,3\}$ with the trivial structure, so that $G=\operatorname{Aut}(A) = \operatorname{Sym}(A)$ is the full symmetric group. For $a=3$, $G(a)$ is the symmetric group on $\{1,2\}$, which is not normal in $G$.

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    @JackSchmidt: Interesting idea, but I'll pass this time. :-)2012-08-15
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No. Let $A=\{0,1,2\}$, with no further structure, so that $G$ is simply $S_3$. The transposition $(12)$ preserves $0$, but $(012)(12)(021)=(02)$, which does not preserve $0$.