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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.

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  • $\begingroup$ Just checking: you're requiring every element $x$ in $G$ to be of finite order? $\endgroup$
    – Todd Trimble
    Commented Jan 16, 2013 at 18:07
  • $\begingroup$ @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. $\endgroup$
    – user30230
    Commented Jan 16, 2013 at 18:19
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    $\begingroup$ For one thing, if $m | n$ and $n \in A$ then $m \in A$. Now you can drop the singleton. $\endgroup$
    – Ramiro de la Vega
    Commented Jan 16, 2013 at 18:22
  • $\begingroup$ @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}. $\endgroup$
    – Todd Trimble
    Commented Jan 16, 2013 at 20:31

2 Answers 2

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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:

H. Deng, M. S. Lucido, W. Shi: The Number of Isomorphism Classes of Finite Groups with Given Element Orders. Algebra and Logic 41 (2002), Issue 1, 39-46.

Andrey Vasil'ev: On finite groups with the given set of element orders. Talk slides, 2010.

V. D. Mazurov: Periodic groups with given element orders. Talk slides, Mal'tsev Meeting, Novosibirsk, November 12-16, 2012.

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For every fixed $n \in \mathbb{N}$, Rolf Brandl and Shi Wujie gave in Finite groups whose element orders 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 $\{1,2,\ldots,i\}$.

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    $\begingroup$ 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 ... $\endgroup$
    – Martin Brandenburg
    Commented Jan 17, 2013 at 1:19
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    $\begingroup$ 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 $n$. $\endgroup$
    – Derek Holt
    Commented Jan 17, 2013 at 9:07
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    $\begingroup$ @Derek: Sorry -- but how does an element of $A_7$ of order 8 look like? $\endgroup$
    – Stefan Kohl ♦
    Commented Jan 17, 2013 at 10:34
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    $\begingroup$ You are right, $A_7$ has spectrum $\{1,2,3,4,5,6,7\}$. Can you think of an example with $\{1,\ldots,8\}$? $\endgroup$
    – Derek Holt
    Commented Jan 17, 2013 at 13:10
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    $\begingroup$ @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 $\mathit{PSL}(3,4)$. $\endgroup$
    – Emil Jeřábek
    Commented Jan 17, 2013 at 13:51

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