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. 



The set of orders of elements in a group 
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](http://link.springer.com/article/10.1023%2FA%3A1014658001689). 
*Algebra and Logic* 41 (2002), Issue 1, 39-46.


Andrey Vasil'ev: [On finite groups with the given set of element orders](http://mat.polsl.pl/groups/docs/Vasilev.pdf).
Talk slides, 2010.


V. D. Mazurov: [Periodic groups with given element orders](http://www.math.nsc.ru/conference/malmeet/12/plenary/mazurov.pdf).
Talk slides, Mal'tsev Meeting, Novosibirsk, November 12-16, 2012.


