Group acting on a set.

Question

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.

Answer 1

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.

Answer 2

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.)