Take the 2-minute tour ×  If the order of all nontrivial
elements in a group is 2, then the group is Abelian. I know of a
proof that is just from calculations (see below). I'm wondering
if there is any theory or motivation behind this fact. Perhaps
to do with commutators?  Proof: $a \cdot b = (b \cdot b) \cdot
(a \cdot b) \cdot (a \cdot a) = b \cdot (b \cdot a) \cdot
(b\cdot a) \cdot a = b \cdot a$.  share|improve this question
It's kind of an odd thing isn't it. Maybe it could help you to
understand how counterexamples exist for groups of exponent p >
2 and how these counterexamples fail when p becomes 2. –  Myself
Jan 11 '13 at 1:19  4 Answers 4  up vote 7 down vote accepted
As every non-identity element has order two, $a^{-1} = a$ for
any element of the group. Therefore $$[a, b] = aba^{-1}b^{-1} =
abab = (ab)^2 = e.$$ Hence the group is abelian. Is this too
calculationy?  share|improve this answer Haha thanks! I knew I
was missing something as obvious as that. I thought the
commutator subgroup was tricky,