Finite groups for which the element orders form an arithmetic progression

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.

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$A_7$ makes a pretty sweet example here. – S. Carnahan Mar 3 '14 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 '14 at 11:04

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.
Is it necessary that the positive integers in the spectrum of such a group be of type 1,2,...,$n$? – Marius Tarnauceanu Mar 3 '14 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 '14 at 14:38