Let $G$ be a group of order $7$ acting on a set of $5$ elements. Show that the action of $G$ must have a fixed point.

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    $\begingroup$ What are your thoughts on this problem? What do you know about group actions? Are there any theorems you think might be useful? $\endgroup$
    – Arthur
    Dec 17, 2013 at 10:47
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    $\begingroup$ What possibilities are there for the size of an orbit? $\endgroup$ Dec 17, 2013 at 10:48
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    $\begingroup$ I do not think that this question should be closed. The OP is interacting with the answers, so is showing effort in their part. This isn't simply a "give me the answer now!" question. $\endgroup$
    – user1729
    Dec 17, 2013 at 12:21
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    $\begingroup$ @YACP: Perhaps Eureka doesn't know about accepting answers? Let's link this page for him/her. And hopefully he/she will accept answers from now on :) $\endgroup$
    – Prism
    Dec 18, 2013 at 2:36
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    $\begingroup$ Thank You. Sorry. I did not noticed it. $\endgroup$
    – EuReka
    Dec 18, 2013 at 12:18

2 Answers 2


The fundamental lemma for a group action is the following. Let $x$ be in $X$, a set on which $G$ acts. There is a map $f: G \rightarrow X$ defined by $g \mapsto g.x$. By definition, the image of this map is the orbit of $x$, denoted by $G.x$. Morever, let $H$ be the subset of $G$ of elements $h$ satisfying $h.x = x$; $H$ is called the stabilizer of $x$ and denoted by $G_x$. This is a subgroup of $G$ and $f$ naturally induces a bijection $\tilde{f}: G/G_x \rightarrow G.x$ (check this).

In particular, we have an equality $|G| = |G_x|.|G.x|$ for any $x$. This shows that the cardinal of any orbit has to divide the cardinal of $G$. You should easily conclude.

  • $\begingroup$ So does this mean that the action must be trivial, with $g\cdot x = x$ for all $g\in G$? $\endgroup$
    – dfeuer
    Dec 17, 2013 at 18:42
  • $\begingroup$ Yes, I'm afraid that the action is not interesting. $\endgroup$ Dec 17, 2013 at 18:46

Hint: The group $S_5$ contains no element of order seven. Conclude that the action is trivial, and so fixes every point.

(Note that, by similar logic, every group of order $49$, or more generally of order $7^n$, must fix a point. See if you can work out why this is.)

  • $\begingroup$ Not getting. Caleys Theorem? $\endgroup$
    – EuReka
    Dec 17, 2013 at 11:53
  • $\begingroup$ @EuReka No, not Caley's Theorem. I am using the fact that an action $G\curvearrowright X$ can be viewed as a homomorphism $G\rightarrow\operatorname{Aut}(X)$ (and vice-versa - homomorphisms can be viewed as actions on an appropriate object, such as a Cayley graph). You may not know this yet though... $\endgroup$
    – user1729
    Dec 17, 2013 at 12:03
  • $\begingroup$ I dont know this .Please explain $\endgroup$
    – EuReka
    Dec 17, 2013 at 12:11
  • $\begingroup$ It is essentially an adaptation of what Jeremy Daniel describes as the "fundamental lemma for a group action". If $G$ acts on a set $X$ then for every element $g\in G$ and every point $x\in X$ you obtain a map $g\mapsto g\cdot x$. Let $g$ be fixed and consider the map $\rho_g: x\mapsto g\cdot x\forall\:x\in X$. Then this is an automorphism of $X$, $\rho_g\in\operatorname{Aut}(X)$. This means that the action corresponds to a map $g\mapsto \rho_g$, and so to a homomorphism $G\rightarrow\operatorname{Aut}(G)$. As required. $\endgroup$
    – user1729
    Dec 17, 2013 at 12:19

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