Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $n$?

share|improve this question
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
add comment

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?

share|improve this answer
Thank you very much. –  Marius Tarnauceanu Oct 1 '11 at 6:34
add comment

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 $\mathcal{C}$?

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.