Order of nontrivial elements is 2 implies Abelian group

Question

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?

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Proof: $a \cdot b = (b \cdot b) \cdot (a \cdot b) \cdot (a \cdot a) = b \cdot (b \cdot a) \cdot (b\cdot a) \cdot a = b \cdot a$.

Answer 1

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?

Answer 2

$[a,b]=1$ for all $a,b\in G$ if and only if $G$ is abelian. You proved that $[a,b]=a^{-1}b^{-1}ab=1$ above - this is the connection to commutators.

I don't know of any strong motivation behind this fact aside from, I guess, knowing that any nonabelian group must have an element of order $>2$. I think that it is just a standard exercise.

It may interest you motivationally to prove that $G/H$ is abelian if and only if $G'\leqslant H$ (if $H \unlhd G$).

Answer 3

Taking inverses reverses the order of multiplication, so if every element is its own inverse multiplication must be commutative.

Answer 4

The idea of this approach is to work with a class of very small, finite, subgroups $H$ of $G$ in which we can prove commutativity. The reason for this is to be able to use the results like Cauchy's theorem and Lagrange's theorem.

Consider the subgroup $H$ generated by two distinct, nonidentity elements $a,b$ in the given group. The group $H$ consists of strings of instances of $a$ and $b$. By induction on the length of a string, one can show that any string of length 4 or longer is equal to a string of length 3 or shorter.

Using this fact we can list the seven possible elements of $H$: $$1,a,b,ab,ba,aba,bab.$$ By (the contrapositive of) Cauchy's Theorem, the only prime divisor of $|H|$ is 2. This implies the order of $H$ is either $1$, $2$, or $4$.

If $|H|=1$ or $2$, then either $a$ or $b$ is the identity, a contradiction.
Hence $|H|$ has four elements. The subgroup generated by $a$ has order 2; its index in $H$ is 2, so it is a normal subgroup. Thus, the left coset $\{b,ba\}$ is the same as the right coset$\{b,ab\}$, and as a result $ab=ba$.