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

Which are the finite groups $G$ such that the element orders of $G$ form an arithmetic progression? Several remarks:

  1. $S_3$, $A_4$ and any $p$-group of exponent $p$ satisfy this property.

  2. If $G$ satisfies this property and $p_1<p_2<...<p_k$ are the prime divisors of $n=\mid G\mid$, then $p_2=2p_1-1$, and consequently $(p_1,p_2)\in \{(2,3),(3,5),(7,13),(19,37),...\}$.

  3. If $G$ satisfies this property and there is $a\in G$ such that $o(a)=\exp(G)$ (in particular, if $G$ is nilpotent), then $G$ is a $p$-group of exponent $p$.

My impression is that the non-nilpotent groups whose element orders form a progression are of order $2^\alpha3^\beta$ (and consequently are solvable), but I failed in proving this.

share|improve this question
1  
$A_7$ makes a pretty sweet example here. –  S. Carnahan Mar 3 at 10:29
    
@S.Carnahan: Yes. -- One can even go one further, i.e. to spectrum $\{1, \dots, 8\}$ -- but that's the maximum. -- I have edited my answer. –  Stefan Kohl Mar 3 at 11:04

1 Answer 1

up vote 9 down vote accepted

No, your impression is not correct. -- A counterexample is the group ${\rm S}_5$, whose elements have orders $1, 2, 3, 4, 5$ and $6$.

As to groups with spectrum (i.e. set of element orders) equal to $\{1, \dots, n\}$: the maximum here is 8, and for $n = 8$ the only such group is an extension of ${\rm PSL}(3,4)$ by a unitary automorphism -- see here.

share|improve this answer
    
Yes, you are right: my impression is not correct. Thanks for your interesting examples! –  Marius Tarnauceanu Mar 3 at 14:06
    
Is it necessary that the positive integers in the spectrum of such a group be of type 1,2,...,$n$? –  Marius Tarnauceanu Mar 3 at 14:15
    
@MariusTarnauceanu: If the spectrum is not of type $1,2,\dots,n$, then there is no element of order 2, hence there are no non-solvable examples in this case. –  Stefan Kohl Mar 3 at 14:38
    
Thanks again! Could I hope to find a classification of solvable groups of odd order satisfying this property? –  Marius Tarnauceanu Mar 4 at 4:54

Your Answer

 
discard

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.