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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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  • $\begingroup$ 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. $\endgroup$ – M Turgeon Aug 15 '12 at 14:30
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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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  • $\begingroup$ I was just writing the same thing. :) $\endgroup$ – tomasz Aug 15 '12 at 14:09
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    $\begingroup$ @tomasz: Makes three of us! $\endgroup$ – Brian M. Scott Aug 15 '12 at 14:10
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    $\begingroup$ We should have 3 similar answers, and let $G$ be the group that exchanges the orders of the answers! $\endgroup$ – Jack Schmidt Aug 15 '12 at 14:10
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    $\begingroup$ @JackSchmidt: Interesting idea, but I'll pass this time. :-) $\endgroup$ – tomasz Aug 15 '12 at 14:12
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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 $0$.

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