# If $|G|<\infty$ acts transitively on a set $X$ with $|X|=10$ then $G$ has an element of order $5$

Let $$G$$ be a finite group that acts transitively on a set $$X$$ with $$|X|=10$$. Show that $$\exists g\in G$$ of order $$5$$

There is a homomorphism $$\phi:G\mapsto S_{10}$$ that sends $$g\in G$$ to its corresponding permutation in $$S_{10}$$.

So $$G$$ is isomorphic to a subgroup of $$S_{10}$$. I want to prove that $$G$$ contains a $$5$$-cycle.

Since $$G$$ acts transitively $$\forall x,y\in \{1,2,...,10\}~\exists g\in G$$ such that $$gx=y.$$ Let $$x$$ be in $$\{1,2,...,10\}$$ Let's consider the set $$\langle g\rangle x =\{x,gx,g^2x,...\}$$. If it contains $$X$$ then $$g$$ is a $$10$$-cycle and $$g^2$$ has order $$5$$.

There must be some cycles in $$G$$ of length $$\ge 3$$ because if we only have disjoint transpositions $$G$$ doesn't act transitively.

I think I need some kind of hint.

This is actually easy. By the orbit-stabilizer theorem $$G$$ has order divisible by $$10$$. By Cauchy's theorem it therefore has an element of order $$5$$.
• @John The strong version of Sylow's first theorem says that there is a subgroup of order $5$, and that is necessarily cyclic. That usually comes up in the proof. If you just use them out of the box, though, Sylow's theorems don't prove it. Cauchy's theorem is very basic and usually proved before Sylow's theorems. – Matt Samuel Jan 6 at 11:40