TITLE: A question on the set of element orders of a finite group
QUESTION [10 upvotes]: Let $G$ be a finite group of order $n$ and denote by $\pi_e(G)$ the set of element orders of $G$. What can be said about $G$ if $\pi_e(G)$ forms a sublattice of the lattice of divisors of $n$?

REPLY [4 votes]: The obvious conjecture following Mark Sapir's post is that $\mathcal{C}$ consists just of the finite nilpotent groups. That is false. Let $P$ be a nonabelian group of order $p^3$ and exponent $p$, for $p$ an odd prime. Then the groups defined by the presentation below have element orders $\{1,2,p,2p\}$ but are not nilpotent.
$$\langle x,y,z,t \mid x^p=y^p=z^p=t^2=(xt)^2=(yt)^2=1, yx=xyz, xz=zx, yz=zy \rangle$$
Further question: are there any non-solvable groups in $\mathcal{C}$?