Welcome to our community

Be a part of something great, join today!

Problem of the Week #288 - Apr 24, 2019

Status
Not open for further replies.
  • Thread starter
  • Moderator
  • #1

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
Here is this week's POTW:

-----
Let $X$ be a finite set with more than one element, and $G$ be a finite group acting transitively on $X$. Show that some element of $G$ is free of fixed points.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Moderator
  • #2

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
This week's problem was solved correctly by castor28 and Olinguito . You can read castor28 's solution below.


Let $|X|=n>1$. As the action is transitive, there is only one orbit. By Burnside's lemma, this is equal to the average number of points fixed by an element of $G$.
The identity of $G$ fixes all the $n$ points. If every element of $G$ fixed at least one point, the average would be grater than $1$.
 
Status
Not open for further replies.