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I have solved the following exercise:

A group of order $55$ acts on a set of order $18$. Then there are at least $2$ fixed points.

But according to my solution, there should be at least 3 fixed points (I solved it via a form of Burnsides lemma and some basic number-theoretic reasoning). Is that true ?

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1 Answer 1

up vote 3 down vote accepted

There might be an orbit of length 11, and one of length 5, so two fixed points are definitely a possibility.

For an example, take the cyclic group generated by the product of two disjoint cycles of length 5 and 11 in the symmetric group $S_{18}$.

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But where is my error then ? My proof went as follows: We know that $\left|X^{G}\right|=\left|X\right|-\sum_{x\in A}\frac{\left|G\right|}{\left|G_{x}\right|}$, where $X^{G}$ denotes the set of all fixed points and $A$ is a system of representatives of the orbits of those points that aren't fixed points. Then we have $\left|X^{G}\right|=18-55\sum\frac{1}{\left|G_{x}\right|}$. Since $18-55\sum\frac{1}{\left|G_{x}\right|}\in\mathbb{N}$ the only options for $\sum\frac{1}{\left|G_{x}\right|}\in\mathbb{Q}$ is to be (...) –  user47574 Jan 28 '13 at 10:44
    
(...) of the form $$ \sum\frac{1}{\left|G_{x}\right|}=\frac{r}{11},\ r\in\left\{ 1,2,3\right\} \ \text{or}\ \sum\frac{1}{\left|G_{x}\right|}=\frac{1}{5}, $$ since otherwise either $55\sum\frac{1}{\left|G_{x}\right|}$ were a purely rational number or $18-55\sum\frac{1}{\left|G_{x}\right|}$ were negative. Summarizing, for the four possible cases above we have $\left|X^{G}\right|\in\left\{ 13,8,3,7\right\} $. –  user47574 Jan 28 '13 at 10:45
    
@user47574 16/55=1/11+1/5. –  peoplepower Jan 28 '13 at 10:56
    
In the example I gave above, you equality reads $2 = 55 - 55/5 - 55/11$. –  Andreas Caranti Jan 28 '13 at 10:57
    
@peoplepower or (Andreas Caranti) Ah, so this is the only case I forgot, right ? (i.e. to fix my proof, there aren't any other cases I have to consider, for which the equation also gives $2$?) –  user47574 Jan 28 '13 at 10:59

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