TITLE: Finite groups with elements of order n
QUESTION [14 upvotes]: 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.

REPLY [6 votes]: This is not really on point, but it may be of some interest. The finite groups of prime exponent are exactly those finite groups that are the set-theoretic union of a collection of pairwise trivially intersecting proper subgroups, all of which have the same order. (See my paper in the Pacific J, of Math. 49, 1973.)