TITLE: Is it true that the order of the group is a power of $2$ if every element has order $2$?
QUESTION [7 upvotes]: I read in this old question that 

If $G$ consists only of elements of order 2, then $|G|=2^m$ for some
   $m$. 

But it's not clear to me. I tested the base case $G=\{a,b,ab,e\}$ but induction does not seem appropriate.

Assuming the statement is true, how does one go about proving this?

REPLY [3 votes]: I would like to propose a more direct answer without using Cauchy's theorem. First note that any group $G$ with only order 2 elements apart form the identity is abelian. Consider the minimum number of generators required to generate the group, say there are $n$ of these. Then every element of the group must be a product of these $n$ generators. Since the group is abelian, the order of the generators within a product is inconsequential and there are at most $\sum_{i=1}^n \binom{n}{i}$ of such products. Adding the identity element itself, this makes at most $2^n$ elements, so $|G|\leq 2^n$. But in fact all of the products must be distinct (and different from the identity) since otherwise we can always rearrange (multiplying by inverse generators and so forth) such that one supposedly independent generator is in fact a product of other generators. Thus $|G|=2^n$, where $n$ is the minimum number of generators of $G$.