Countable structures with uncountable many automorphisms - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T21:30:29Z http://mathoverflow.net/feeds/question/103807 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103807/countable-structures-with-uncountable-many-automorphisms Countable structures with uncountable many automorphisms Ioannis Souldatos 2012-08-02T18:38:19Z 2012-08-02T18:50:35Z <p>The following is supposed to be "clear" according to Kueker, but I could not see why. Can anyone help?</p> <p>Let $A$ be a countable structure with uncountable many automorphisms. Then for every $\vec{a}\in A^{&lt;\omega}$, $(A,\vec{a})$ has a non-trivial automorphism, i.e. there exists some $f:A\rightarrow A$ such that $f\neq id$ and $f(a)=a$, for all $a\in \vec{a}$. </p> <p>Note: The argument is part of a larger proof from <a href="http://www.zentralblatt-math.org/portal/en/zmath/search/?q=an:0316.02018&amp;format=complete" rel="nofollow">Infinitary Logic: In memoriam Carol Karp</a> </p> http://mathoverflow.net/questions/103807/countable-structures-with-uncountable-many-automorphisms/103809#103809 Answer by Trevor Wilson for Countable structures with uncountable many automorphisms Trevor Wilson 2012-08-02T18:43:38Z 2012-08-02T18:50:35Z <p>Let $\vec{a} \in A^{&lt;\omega}$. There are uncountably many automorphisms $f$ and only countably many possible values for $f(\vec{a})$, so there must be two different automorphisms $f_1$ and $f_2$ with $f_1(\vec{a}) = f_2(\vec{a})$. Then $f_2^{-1} \circ f_1$ is a nontrivial automorphism fixing $\vec{a}$.</p> <p>In fact, there are uncountably many automorphisms moving $\vec{a}$ the same way, so fixing one and composing the others with its inverse gives uncountably many nontrivial automorphisms fixing $\vec{a}$.</p>