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Let $A$ be a subset of natural numbers. Consider the following problem:

Is there a group $G$ such that $\lbrace O(x) \; | \; x \in G \rbrace = A\cup\lbrace 1\rbrace$ ? (where $O(x)$ is the order of $x$)

If you know any reference concerning this problem or any partial solution (containing a necessary or sufficient condition) , please let me know.

flag	asked Jan 16 at 17:55 chatish 2,193●4●20

Just checking: you're requiring every element $x$ in $G$ to be of finite order? – Todd Trimble Jan 16 at 18:07

@Todd: Yes. Note that this does not affect the generality of the problem. since if $B = A\cup\lbrace\infty\rbrace$ then $G\oplus\mathbb{Z}$ whould be the answer. – chatish Jan 16 at 18:19

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For one thing, if $m | n$ and $n \in A$ then $m \in A$. Now you can drop the singleton. – Ramiro de la Vega Jan 16 at 18:22

@shatich: it does affect the problem a little, in that the left side of your equation does not read {O($x$): $x \in G$ and $x$ is of finite order}. – Todd Trimble Jan 16 at 20:31

Stefan Kohl · Accepted Answer · 2013-01-16 18:27:45Z UTC

Obviously not for every set $A \subset \mathbb{N}$ there is a group $G$ with $A$ as set of orders of its elements (usually called 'spectrum') -- for example if $G$ has an element of order $n$, then $G$ also has an element of order $d$ for every divisor $d$ of $n$.

For a survey of what is known on this question, you may check the following references: http://link.springer.com/article/10.1023%2FA%3A1014658001689?LI=true, http://mat.polsl.pl/groups/docs/Vasilev.pdf, http://www.math.nsc.ru/conference/malmeet/12/plenary/mazurov.pdf.

J. H. S. · Answer 2 · 2013-01-17 08:17:04Z UTC

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For every fixed $n \in \mathbb{N}$, Rolf Brandl and Shi Wujie gave in Finite groups whose elements are consecutive integers (Journal of Algebra, 143, 388-400 (1991).) a complete classification of finite groups whose spectrum is $\{1,2,\ldots,n\}$. A particularly appealing spin-off of their study is the following one:

Let $i$ be a positive integer greater than $8$. There is no finite group $G$ whose spectrum is $\{1,2,\ldots,i\}$.

edited Jan 17 at 8:17

answered Jan 16 at 18:37

J. H. S.
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I like these results about quite general and abstract questions where suddenly such an explicit bound pops up. For example the automorphism group of a smooth algebraic curve of genus $g>1$ has order at most ${\bf 84}(g-1)$. It always makes me wonder if there is any deeper reason behind this. For example, your answer says that there groups with spectrum $\{1,\dotsc,8\}$, but the spectrum $\{1,\dotsc,9\}$ is not possible. Why on earth ... – Martin Brandenburg Jan 17 at 1:19

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Since the proof of this result involves considering all finite simple groups using the classification, it is unlikely that there is any easily described reason for this result. Note that the group $A_7$ has spectrum $\{1,\ldots,8\}$. As $n$ gets larger, it gets harder to avoid commuting elements resulting in elements of order higher than $n$. – Derek Holt Jan 17 at 9:07

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@Derek: Sorry -- but how does an element of $A_7$ of order 8 look like? – Stefan Kohl Jan 17 at 10:34

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@Derek: According to the Brandl and Wujie paper, every such group is isomorphic to $[\mathit{PSL}(3,4)]\langle\beta\rangle$, where $\beta$ is a unitary automorphism of $\mathit{PSL}(3,4)$. – Emil Jeřábek Jan 17 at 13:51

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Thanks Emil! So there is a unique such group. – Derek Holt Jan 17 at 14:11

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The set of orders of elements in a group

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