A group $G$ of order $32$ act in a set $X$ order $15$.Show that there is at least one element in set $X$ that remains stable under the action of $G$

Question

A group $G$ of order $32$ act in a set $X$ order $15$.Show that there is at least one element in set $X$ that remains stable under the action of $G$

Any ideas and hints to show this?

Answer 1

Since $|G|=32 = 2^5$, every element of $G$ must have a power of $2$ as its order. Consider the isotropy group $$G(x) = \{g\in G : g(x)= x\}$$ This is a subgroup of $G$, for if $g,h\in G(x)$ then $g(h(x))=g(x)=x$. If we let $G_x$ denote the orbit of $x\in X$ then we have $|G_x|\cdot |G(x)| = |G|$. See http://mathworld.wolfram.com/GroupOrbit.html and http://mathworld.wolfram.com/IsotropyGroup.html for more information.

Since $|G|$ is a power of $2$, then also $|G_x|$ and $|G(x)|$ must be powers of 2. This is true for every $x\in X$, and the orders of the orbits must sum to 15, for there are 15 elements in total. The only way for some powers of 2 to sum to 15 is if there is at least one 1 (ie. $2^0$).

Answer 2

$X$ is partitioned into its orbits under $G$. The class formula shows the cardinal of an orbit is a divisor of $\lvert G\rvert$, hence a power of $2$.If there were no fixed point, the cardinal of each orbit would be a positive power of $2$, and $\lvert X\rvert$ would be even.