Let $G$ a nilpotent group such that $10$ divides $|G|$. It is true that $G$ has element of order $10$?

Question

Let $G$ a nilpotent group such that $10$ divides $|G|$. It is true that $G$ has element of order $10$?

We know that $G$ can be expressed like direct sum of Sylow's subgroups and that $G$ has a normal subgroup of order $10$ but how can we answer to this question?

Answer 1

Since $G$ is a direct product of its Sylow subgroups, let us call its $2-$ and $5-$ Sylow subgroups $P_2$ and $P_5$. They are non-trivial since $|G|$ is divisible by $10$. Taking a non-identity element of $P_2$ and raising to an appropriate power, we have $g\in P_2$ with $o(g)=2$ and similarly we have $h \in P_5$ with $o(h)=5$. But since $G = P_2 \times P_5 \times \mathrm{Stuff} $, all elements of $P_2$ commute with all elements of $P_5$, so $g$ and $h$ commute. Therefore, $o(gh)=10$, so the answer is affirmative and $gh$ is the required element.

Answer 2

There are elements of order $5$ and $2$, by Cauchy.

If I can show those elements commute, then their product will have order $10$.

But nilpotency and finiteness imply that elements of relatively prime order commute.