Showing that if all nonidentity elements have same order, the group is elementary abelian. I can see that if $N$ is a group such that all $g \ne 1, g \in N$ have the same order, then this order is some prime $p$.   Why is $N$ elementary abelian and of order $p^m$ for some $m$?
 A: This is true for $p = 2$. Let $a$ and $b$ two elements, then $(ab)^2 = (ba)^2 = e$ which means $abab = baba$. 
Then, $ab(ab ab) = ba b(a a)b \Rightarrow ab = ba(bb) = ba$.
A: As Tobias Kildetoft mentioned in his comment, this need not be true. However, it is true if we assume that $N$ is abelian.
Let $N$ be a finite group such that 


*

*$N$ is abelian

*all elements of $N$ have the same order.


Just from premise 2. we can conclude that $N$ has prime power order. For suppose $p$ and $q$ were two distinct primes such that $p\mid |N|$ and $q\mid |N|$. Then Cauchy's theorem implies that there are elements of order $p$ and $q$ contradicting 2. Thus there can be at most one prime dividing the order of $N$.
Assuming $N$ is abelian, we can conclude from the fundamental theorem of finite abelian groups (to use a sledgehammer) that $N$ is isomorphic to a direct product of cyclic groups with prime power order. If any of the factor groups were a cyclic group of order $p^i$ for $i$ greater than 1, we would know that there was an element of order $p^i$. This contradicts 2. as Cauchy's theorem still implies that there is an element of order $p$.
