Consider a finite group where all elements have the same order $n$.
What could be said about such groups?


For $n=2$ it could be proved that such group is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^k$.
Could it be somehow generalized on case $n>2$?


**EDIT:** Surely the identity has order 1, so we have to exclude it.



Finite groups with elements of order n
This question is closely related to the [restricted Burnside problem](http://en.wikipedia.org/wiki/Restricted_Burnside_problem): given numbers $m$ and $p$, is the restricted Burnside group $B\_0(m,p)$ finite? Every group with $m$ generators of exponent $p$ is the quotient of the Burnside group $B(m,p)=F\_m/\langle w^p\rangle,$ where $F\_m$ is a free group with $m$ generators, and $B\_0(m,p)$ is the quotient of $B(m,p)$ by the intersection of all subgroups of finite index (which is a normal subgroup). For the case of *prime* exponent, A.I. Kostrikin proved that the restricted Burnside problem has affirmative solution (and Efim Zelmanov proved it in general). Thus the answer to the original question is:



> 
> A finite group $G$ has the property that all non-unit elements have the same order $p$ if and only if $p$ is prime and $G\ne 1$ is a quotient of $B\_0(m,p)$ for some $m.$ 
> 


For small values of $p$, even the Burnside group $B(m,p),$ which is somewhat easier to study, is known to be finite ($p=2,3$) and one may hope to get a more precise answer (for $p=2$ the group is elementary abelian 2-group of rank $m$). 


