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

Question

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.

Answer 1

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 $5$.