The set of orders of elements in a group  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. 
 A: 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 $\{1,2,\ldots,i\}$. 
A: 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.
