Does there exist a finite group with orders of elements $1$, $2$, ..., $8$ (but no elements of larger order)?

For which $n$ does there exist a finite group with orders of elements $1$, $2$, $\ldots$, $n$ (but without elements of larger order)?

I could find such groups for $n=1,2,\ldots,7$:

• For $n=1,2,3,4$ we can take the symmetric group $S_n$;
• For $n=5$ we can take the alternating group $A_6$;
• For $n=6$ we can take the symmetric group $S_5$ (or $S_6$);
• For $n=7$ we can take the alternating group $A_7$.

Does such a group also exist for larger values of $n$?

By Lagrange's theorem we know that the order of the group must be a multiple of the lcm of $1$, $2$, $\ldots$, $n$. Furthermore, the group cannot be abelian since in an abelian group containing elements of order $a$ and $b$ there is also an element of order $\text{lcm}(a,b)$.

Solution 1:

The set of orders of the elements of a group $G$ is called the spectrum of $G$. A complete classification of all finite groups with spectrum $(1,\dots,n)$ was made by Rolf Brandl and Shi Wujie in Finite groups whose element orders are consecutive integers (Journal of Algebra, Volume 143, Issue 2, November 1991, Pages 388-400).

According to this article the answer to your question is yes; there is one unique finite group with spectrum $(1,...,8)$, namely $PSL(3,4)\langle\beta\rangle$ where $\beta$ is a unitary automorphism.

$8$ is however the largest $n$ for which there exists a finite group with spectrum $(1,...,n)$.

Another interesting fact is that $A_7$ is the only finite group with spectrum $(1,...,7)$.

A related answer on Mathoverflow can be found here.