Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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 $S_{18}$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.