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 $n$?

Let $G$ be a finite group, $n(G)$ the l.c.m. of orders of elements in $G$. Here are some obvious observations. A group $G$ belongs to your class $\mathcal C$ iff $G$ contains the cyclic group of order $n(G)$. Every $p$group belongs to $\mathcal C$. The class is closed under direct products of groups with coprime orders. Hence it contains all finite nilpotent groups. Hence the class is closed under all (finite) direct products. For every group $G$, the group $G\times {\mathbb Z}/n(G){\mathbb Z}$ is in $\mathcal C$. What else do you want to know? 


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 nonsolvable groups in $\mathcal{C}$? 

