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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 $n$?

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I took the liberty of adding the "finite-groups" tag, hope you don't mind. –  Yemon Choi Sep 29 '11 at 5:38
    
It would also be interesting to know some nontrivial classes of groups which satisfy this. –  Gjergji Zaimi Sep 29 '11 at 10:16
    
It doesn't directly answer your question, but the following paper gives some idea of how deep such questions about element orders can be: www.mathematik.uni-kl.de/~malle/download/elemords.pdf –  Colin Reid Sep 29 '11 at 13:07

2 Answers 2

up vote 13 down vote accepted

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 co-prime 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?

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Thank you very much. –  Marius Tarnauceanu Oct 1 '11 at 6:34

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 $\mathcal{C}$?

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Derek: I wrote that for every finite group $G$ the group $G\times {\mathbb Z}/n(G){\mathbb Z}$ is in $\mathcal C$. So every finite group embeds into a group from $\mathcal C$. –  Mark Sapir Sep 30 '11 at 10:18
    
Mark: Sorry, I missed that! –  Derek Holt Sep 30 '11 at 10:25
    
It's a very interesting example. Thank you very much. I was thinking to another example: non-nilpotent dihedral groups $D_{2n}$ with $n$ even. –  Marius Tarnauceanu Oct 1 '11 at 6:40

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